Let ABCD be a quadrilateral.
Go round the quadrilateral in order: A → B → C → D → back to A. The sides are AB, BC, CD, DA.
(i) Two sides are adjacent when they share an endpoint. AB has endpoints A and B: the other side at B is BC and the other side at A is DA. So the sides adjacent to AB are BC and DA. The only side sharing no endpoint with AB is CD — that is the side opposite to AB.
(ii) Angles at the two ends of a side are adjacent. \(\angle A\) is joined to B by side AB and to D by side DA, so the angles adjacent to \(\angle A\) are \(\angle B\) and \(\angle D\). The only remaining vertex is C, so the angle opposite to \(\angle A\) is \(\angle C\).
(iii) Opposite sides are two sides of the quadrilateral that have no endpoint in common. Opposite angles are the internal angles at two vertices that are not the two ends of any side (equivalently, the two ends of a diagonal).
* You have used internal angles of quadrilaterals, but they too require an exact definition, just like how we gave one for a quadrilateral. Precisely define the internal angle of a quadrilateral at a given vertex. Your answer should work for a non-convex quadrilateral too. (Hint: use the opposite vertex as well.)
Take quadrilateral ABCD and the vertex A. The two sides at A are AB and AD. The rays AB and AD make two angles that add up to \(360^\circ\): the ordinary angle \(\angle BAD\) (less than \(180^\circ\)) and the reflex angle \(360^\circ - \angle BAD\). The question is which of the two lies inside the quadrilateral.
Now use the opposite vertex C. The other three vertices B, C, D form a triangle \(\triangle BCD\).
Definition. The internal angle of ABCD at A is
Why this is right: if A is outside \(\triangle BCD\), the quadrilateral is the triangle-like shape bulging out at A, so the inside is the ordinary angle. If A is inside \(\triangle BCD\), then A is a dent — the quadrilateral is \(\triangle BCD\) with the small triangle near A bitten out (like D in DART), so the inside of the figure at A is the reflex angle.
The same rule applies at B, C and D, each time using the triangle formed by the other three vertices. A convex quadrilateral has no vertex inside the triangle of the other three, so all four internal angles are less than \(180^\circ\), as expected.
In a quadrilateral ABCD, suppose \(AB \parallel DC\). Can ABCD be non-convex? What if we instead assume \(AB = CD\)? What if we instead assume \(\angle A = \angle C\)?
(a) \(AB \parallel DC\): No, ABCD must be convex.
Because ABCD does not cross itself, B and C lie on the same side of line AD, so AB and DC point the same way. Then AD is a transversal of the parallel lines AB and DC, and \(\angle A\) and \(\angle D\) are co-interior angles:
In the same way BC is a transversal, so
Each internal angle is part of a pair adding to \(180^\circ\), so none of them can be more than \(180^\circ\). All four are less than \(180^\circ\), so ABCD is convex. (If AB and DC pointed opposite ways, sides AD and BC would cross — that is a self-intersecting 4-gon, not a quadrilateral.)
(b) \(AB = CD\): Yes, it can be non-convex. Example: A(0, 0), B(4, 0), C(1, 1), D(1, 5).
Here C is a dent: C lies inside \(\triangle ABD\), so the internal angle at C is more than \(180^\circ\).
(c) \(\angle A = \angle C\): Yes, it can be non-convex. Example: A(−3, 0), B(0, 1), C(3, 0), D(0, 5). This arrow-head is symmetric about the line BD, so \(\angle A = \angle C\), and B is a dent (B lies inside \(\triangle ACD\)), so the angle at B is reflex.
Consider three non-collinear points A, B, C and draw the lines AB, BC, CA. For every possible location of point D in the plane outside these lines, decide if ABCD is self-intersecting, non-convex, or convex. (Hint: the three lines divide the plane into 7 regions.)
Remember that ABCD means the sides AB, BC, CD, DA, so D is joined to C and to A. The three lines cut the plane into 7 regions (see the figure): the inside of the triangle, three regions across a side, and three regions beyond a corner.
| Where D is | ABCD is ... because |
|---|---|
| Inside \(\triangle ABC\) | non-convex: D lies inside \(\triangle ABC\), so D is a dent. |
| Across side CA (opposite B) | convex: B and D are on opposite sides of AC and the diagonals AC, BD cross. |
| Across side AB (opposite C) | self-intersecting: side CD must cross side AB to reach D. |
| Across side BC (opposite A) | self-intersecting: side DA must cross side BC to reach A. |
| Beyond corner A | non-convex: A lies inside \(\triangle BCD\), a dent at A. |
| Beyond corner B | non-convex: B lies inside \(\triangle ACD\), a dent at B. |
| Beyond corner C | non-convex: C lies inside \(\triangle ABD\), a dent at C. |
So ABCD is convex in exactly one region (across CA, the side not used by ABCD), self-intersecting in two regions, and non-convex in the other four.
Can a quadrilateral be both self-intersecting and non-planar?
No. Suppose two sides cross, say AB and CD meet at a point E.
Two lines that meet at a point lie in one plane. So line AB and line CD lie in a single plane. That plane contains A and B (they are on line AB) and C and D (they are on line CD).
So all four vertices are in one plane and the 4-gon is planar. A self-intersecting 4-gon is therefore always planar; a non-planar 4-gon can never cross itself.
True or false?
(i) True. Adjacent angles of a parallelogram add to \(180^\circ\) and opposite angles are equal (Theorem 1). If \(\angle A = 90^\circ\):
All four angles are right angles, so it is a rectangle.
(ii) False. Every rhombus has perpendicular diagonals, so this condition adds nothing. Counterexample: A(0, 0), B(5, 0), C(8, 4), D(3, 4). All sides are 5 and the diagonals AC and BD are perpendicular, but \(\angle A\) is not \(90^\circ\), so it is not a square.
(iii) True. Let ABCD be a parallelogram with \(AC = BD\). In \(\triangle ABC\) and \(\triangle BAD\):
So \(\triangle ABC \cong \triangle BAD\) (SSS), which gives \(\angle B = \angle A\). But \(\angle A + \angle B = 180^\circ\), so
A parallelogram with a right angle is a rectangle, by (i).
The diagonal AC of a parallelogram ABCD bisects \(\angle A\). Show that it also bisects \(\angle C\) and that ABCD is a rhombus.
Write \(\angle BAC = \angle CAD = x\) (AC bisects \(\angle A\)).
AC bisects \(\angle C\). AB \(\parallel\) DC with transversal AC gives alternate angles, and AD \(\parallel\) BC with transversal AC gives alternate angles:
So \(\angle DCA = \angle BCA\): AC bisects \(\angle C\).
ABCD is a rhombus. In \(\triangle ABC\), \(\angle BAC = x = \angle BCA\), so the sides opposite these equal angles are equal:
Opposite sides of a parallelogram are equal, so \(AB = CD\) and \(BC = DA\). Hence
All four sides are equal, so ABCD is a rhombus.
The following questions examine converses of true properties. Answer them with Yes or No. If your answer is No, what extra condition can you add so that the answer becomes Yes?
(i) Yes. AC bisects \(\angle A\) and \(\angle C\). Compare \(\triangle ABC\) and \(\triangle ADC\):
So \(\triangle ABC \cong \triangle ADC\) (ASA), giving \(AB = AD\) and \(CB = CD\). In the same way BD bisects \(\angle B\) and \(\angle D\), so \(\triangle BAD \cong \triangle BCD\), giving \(BA = BC\) and \(DA = DC\). Together:
so ABCD is a rhombus.
(ii) Yes. Diagonals that bisect each other make ABCD a parallelogram (Theorem 4). Let them meet at O. In \(\triangle AOB\) and \(\triangle AOD\):
So \(\triangle AOB \cong \triangle AOD\) (SAS) and \(AB = AD\). A parallelogram with two adjacent sides equal has all four sides equal, so it is a rhombus.
(iii) No. An isosceles trapezium has equal diagonals but is not a rectangle. Example: A(0, 0), B(6, 0), C(4, 3), D(2, 3):
so \(AC = BD = 5\), yet \(\angle A\) is not a right angle.
Extra condition: the diagonals also bisect each other. Then ABCD is a parallelogram (Theorem 4) with equal diagonals, which is a rectangle (Exercise Set 12.2, Q1 (iii)).
Let ABCD be a parallelogram with \(AB \ne BC\). Show that the pairwise intersection points of the four angle bisectors form the vertices of a rectangle (see Fig. 12.11). Why did we assume \(AB \ne BC\)?
Name the points: the bisectors of \(\angle A\) and \(\angle B\) meet at R, of \(\angle B\) and \(\angle C\) at Q, of \(\angle C\) and \(\angle D\) at P, of \(\angle D\) and \(\angle A\) at S.
Each corner of PQRS is a right angle. Look at \(\triangle ARB\). Since AD \(\parallel\) BC, the adjacent angles add up to \(180^\circ\):
The bisectors take half of each:
So the third angle of \(\triangle ARB\) is
\(\angle SRQ\) is formed by the same two lines (the bisectors from A and B) as \(\angle ARB\), so it is either the same angle or vertically opposite to it; either way \(\angle SRQ = 90^\circ\). Exactly the same argument with \(\angle B + \angle C\), \(\angle C + \angle D\) and \(\angle D + \angle A\) gives right angles at Q, P and S.
A quadrilateral with four right angles is a rectangle, so PQRS is a rectangle.
Why \(AB \ne BC\)? If \(AB = BC\), ABCD is a rhombus. Then every diagonal bisects the angles at its ends (Exercise Set 12.2, Q2), so all four bisectors pass through the centre of the rhombus and P, Q, R, S are the same point. There is no rectangle — it shrinks to a single point.
(i) By the Midpoint Theorem each midpoint segment is half the side it does not touch:
The midpoints also halve the sides:
Now list each triangle's sides in the order (half of BC, half of AB, half of AC):
Matching sides are equal, so the triangles are congruent by SSS, with the vertices matched in the order written.
So \(\triangle PQR \cong \triangle QPA \cong \triangle RBP \cong \triangle CRQ\): the four small triangles are all congruent, which is what the paper-cutting experiment showed.
(ii) Yes. Through each vertex of \(\triangle PQR\), draw the line parallel to the opposite side: through P parallel to QR, through Q parallel to PR, through R parallel to PQ. These three lines meet in pairs at A, B and C.
Why it works: in the original figure \(QR \parallel AB\) (Midpoint Theorem), so line AB is the line through P parallel to QR. Likewise line AC is the line through Q parallel to PR, and line BC is the line through R parallel to PQ. Two of these lines meet at exactly one point, so the vertices come back exactly. (APRQ, PBRQ and PRCQ are parallelograms, so for instance A is the fourth corner of the parallelogram on PR and RQ.)
In \(\triangle ABC\), let M and N be midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.
Let MN meet AD at X. We show X is the midpoint of AD.
By the Midpoint Theorem in \(\triangle ABC\), \(MN \parallel BC\). So the line MX is parallel to BD.
Now look at \(\triangle ABD\). M is the midpoint of AB, and the line through M parallel to BD meets AD at X. By Theorem 7 (converse of the Midpoint Theorem) this line bisects the third side AD:
So MN passes through the midpoint of AD, i.e. MN bisects AD, wherever D is on BC.
In a quadrilateral ABCD, suppose \(AB \parallel DC\) and \(AB \ne CD\). Suppose G and H are the midpoints of AC and BD respectively. Prove that \(GH \parallel AB\). (Why did we assume \(AB \ne CD\)?)
Let M be the midpoint of AD.
In \(\triangle ADC\): M and G are the midpoints of AD and AC, so by the Midpoint Theorem
In \(\triangle DAB\): M and H are the midpoints of DA and DB, so
Since \(AB \parallel DC\), the lines MG and MH are both parallel to AB and both pass through M. There is only one line through M parallel to AB, so M, G, H all lie on that line. Hence line GH is that line, and
Why \(AB \ne CD\)? If also \(AB = CD\), then ABCD is a parallelogram (Theorem 5). Its diagonals bisect each other, so G and H are the same point. Then there is no segment GH at all, and the statement has no meaning.
Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively.
(i) By Theorem 9 (the Midpoint Theorem for Quadrilaterals), PQRS is a parallelogram. PR and QS are its diagonals, and the diagonals of a parallelogram bisect each other (Theorem 1(c)). So PR and QS bisect each other.
(ii) Apply the Midpoint Theorem in \(\triangle ABC\) and in \(\triangle BCD\):
If \(AC = BD\) then \(PQ = QR\). So the parallelogram PQRS has two adjacent sides equal, i.e. it is a rhombus. The diagonals of a rhombus are perpendicular, so \(PR \perp QS\).
The converse is true too. If \(PR \perp QS\), then PQRS is a parallelogram whose diagonals bisect each other at right angles, so it is a rhombus (Exercise Set 12.2, Q3 (ii)). Then \(PQ = QR\), and
Suppose PQRS is the Varignon parallelogram of ABCD.
(i) PQRS alone does not fix ABCD: many different quadrilaterals share the same Varignon parallelogram. So one extra piece of information must be carried over — the position of one vertex. Copy the position of A relative to PQRS (for example, copy the lengths PA and SA, which fixes A as the corner of a triangle on PS). Call this point A′. Then:
• extend A′P beyond P by its own length to get B′ (so P is the midpoint of A′B′);
• extend B′Q by its own length to get C′;
• extend C′R by its own length to get D′.
(ii) Why S comes out right. Let S′ be the midpoint of D′A′. By Theorem 9, the midpoints P, Q, R, S′ of A′B′C′D′ form a parallelogram PQRS′. But PQRS is also a parallelogram, and a parallelogram with three given corners P, Q, R (in that order) has only one possible fourth corner. So \(S' = S\): S lies on A′D′ and is its midpoint. The construction always closes up.
Why A′B′C′D′ is congruent to ABCD. A′ was placed by copying PA and SA, so \(\triangle A'PS \cong \triangle APS\) (SSS): the copied figure PQRS-with-A′ is an exact copy of the original PQRS-with-A. In the original, B is found from A by the very same rule we used for B′ (P is the midpoint of AB), and likewise C from B and D from C. Doing the same constructions on two exact copies gives exact copies, so B′, C′, D′ sit where B, C, D sat, and \(A'B'C'D' \cong ABCD\). (For instance \(\triangle SDR \cong \triangle SD'R\) because the copy of D is constructed from the copy of the same figure.)
The \"multiple ways\": you could start from any vertex (place B′, C′ or D′ first) or go round the other way; every choice gives the same quadrilateral once that first vertex is placed correctly. (If A′ is put somewhere else, the construction still closes up, but it makes a different quadrilateral with the same Varignon parallelogram.)
(iii) By the Midpoint Theorem in \(\triangle ABC\) and \(\triangle BCD\):
If PQRS is a square, \(PQ = QR\) gives \(AC = BD\), and \(PQ \perp QR\) gives \(AC \perp BD\) (lines parallel to two perpendicular lines are perpendicular).
Converse. If \(AC = BD\) and \(AC \perp BD\), then \(PQ = QR\) and \(PQ \perp QR\). PQRS is a parallelogram (Theorem 9) with a right angle (so a rectangle) and two adjacent sides equal (so a rhombus): it is a square.
Justify why the plane cannot be tiled with a regular pentagon. (Hint: Read the first 3 sentences of ‘Think and Reflect’ in the section on tiling.) (There are many ways to tile the plane using a suitable irregular pentagon. The most recent method was found in 2015.)
The angles of a pentagon add up to \((5 - 2) \times 180^\circ = 540^\circ\), so each angle of a regular pentagon is
In a tiling, look at any point where corners of tiles meet. The angles around that point must add up to exactly \(360^\circ\). Around the point there are only pentagon corners, each \(108^\circ\):
So no whole number of corners fits. A corner could also touch the middle of another tile's side, where the straight side already uses \(180^\circ\) and the corners must fill the other \(180^\circ\); but
and neither is \(180^\circ\). So regular pentagons always leave gaps or overlap: they cannot tile the plane.
Draw a non-convex 4-gon DART. Show how we can tile the plane with copies of DART. Both methods that we discussed earlier will work. Which do you prefer?
Label the angles of DART 1, 2, 3, 4 (one of them, at the dent, is more than \(180^\circ\)). They still add up to \(360^\circ\).
Method 1 (half-turns). Take a copy of DART and turn it through \(180^\circ\) about the midpoint of one side, say AR. The copy fits exactly along AR (the half-turn swaps A and R). Repeat at the midpoint of every side of every copy you add. Around each vertex you get the four angles 1, 2, 3, 4 once each, which add to \(360^\circ\), so the copies close up with no gaps and no overlaps — even the reflex angle fits, because the other three angles together make up the rest of \(360^\circ\).
Method 2 (Varignon grid). Join the midpoints of DART's sides to get its Varignon parallelogram. Tile the plane with that parallelogram (a grid of parallel lines, as in the introduction). Then place a copy of DART on every other parallelogram of the grid, exactly as the green copies sit in Fig. 12.32; the gaps that are left are copies of DART turned through \(180^\circ\).
Which to prefer? This is your choice. Method 1 is easier to do with paper cut-outs: you only ever rotate about a midpoint. Method 2 is easier to draw on paper: once the grid of parallel lines is drawn, every copy's position is fixed by the grid.
Using a fact about parallelograms, show how to tile the plane using any given triangle. (Hint: Can you use the parallelogram tiling in the introduction?)
Take \(\triangle ABC\) and a second copy turned through \(180^\circ\) about the midpoint M of BC. The half-turn swaps B and C and sends A to a point D with M the midpoint of AD.
Now the diagonals AD and BC of ABDC bisect each other at M, so ABDC is a parallelogram (Theorem 4). It is made of exactly two copies of the triangle.
The introduction showed that the plane can be tiled by copies of any parallelogram (a grid of two sets of parallel lines). Tile the plane with copies of ABDC, then cut each parallelogram along its diagonal BC. Every tile becomes two copies of \(\triangle ABC\), so the plane is tiled by the triangle.
Mark the midpoint of the line drawn on the paper (see Fig. 12.33), given that the horizontal lines are equally spaced. Justify your answer.
Number the ruled lines 1 to 7 from the top. The drawn segment XY starts on line 6 and ends on line 2, so it crosses four equal gaps. Its midpoint M is the point where it crosses line 4, the middle line.
Justification. Call the crossing points on lines 6, 5, 4, 3, 2 \(X = X_6, X_5, X_4, X_3, X_2 = Y\). Take any three neighbouring lines, say k−1, k, k+1, and draw a second crossing line, perpendicular to the ruled lines, meeting them at \(Z_{k-1}, Z_k, Z_{k+1}\). Because the lines are equally spaced, \(Z_k\) is the midpoint of \(Z_{k-1}Z_{k+1}\).
Now \(X_{k-1}X_{k+1}Z_{k+1}Z_{k-1}\) is a trapezium (two of its sides lie on the parallel lines k−1 and k+1). Line k passes through the midpoint \(Z_k\) of one slanting side and is parallel to the parallel sides, so by End-of-Chapter Q7 (i) it meets the other slanting side at its midpoint: \(X_k\) is the midpoint of \(X_{k-1}X_{k+1}\). So neighbouring pieces are equal:
Two pieces lie on each side of \(X_4\), so \(X_4\) is the midpoint M of XY.
You know that the sum of angles of a quadrilateral is \(360^\circ\), even for a non-convex quadrilateral. (Recall the proof.) Now consider a self-intersecting quadrilateral ABCD, where AB and CD intersect at point E. Show that \(\angle A + \angle B + \angle C + \angle D < 360^\circ\). Can you construct ABCD such that \(\angle A + \angle B + \angle C + \angle D = 2^\circ\)?
The angles of ABCD are \(\angle DAB\), \(\angle ABC\), \(\angle BCD\), \(\angle CDA\). Because E lies on AB and on CD, each of them is an angle of one of the two triangles \(\triangle EBC\) and \(\triangle EAD\):
\(\angle AED = \angle BEC\) (vertically opposite angles). Adding:
Since \(\angle BEC > 0^\circ\), the sum is less than \(360^\circ\).
A sum of \(2^\circ\). We need
Construction: draw two lines crossing at E at an angle of \(179^\circ\) (i.e. \(1^\circ\) apart). On one line mark A and B on opposite sides of E; on the other line mark C and D on opposite sides of E, with C in the \(179^\circ\) angle next to B. Then AB and CD cross at E with \(\angle BEC = 179^\circ\), and the angle sum of ABCD is exactly \(2^\circ\). (The quadrilateral is a very thin bow-tie.)
In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that \(DP = BQ\) (see Fig. 12.34). Show that APCQ is a parallelogram.
Let the diagonals AC and BD of ABCD meet at O. Diagonals of a parallelogram bisect each other (Theorem 1(c)):
P and Q are on BD with \(DP = BQ\), so subtracting equal lengths from equal lengths:
So the diagonals AC and PQ of APCQ meet at O and bisect each other: \(OA = OC\) and \(OP = OQ\). By Theorem 4, APCQ is a parallelogram.
A right-triangle shaped cutout of a paper is folded such that point A touches point B (Fig. 12.35). Show that the crease line can be used to find the midpoint of not only AB but also that of AC.
When A is folded onto B, every point of the crease is the same distance from A and from B. So the crease is the perpendicular bisector of AB: it passes through the midpoint M of AB and is perpendicular to AB. That marks the midpoint of AB.
The triangle is right-angled at B, so \(BC \perp AB\). The crease and BC are both perpendicular to AB, hence
In \(\triangle ABC\), the crease is the line through the midpoint M of AB parallel to BC. By Theorem 7 (converse of the Midpoint Theorem) it bisects AC. So the point N where the crease meets AC is the midpoint of AC.
You saw how to use the Midpoint Theorem to divide a given triangle into 4 congruent triangles. Can we divide a triangle into 3 congruent triangles? This exercise shows us how to do that and more, provided we are allowed to cut and reassemble.
(i) Turn the piece \(\triangle MPB\) through \(180^\circ\) about P. Since P is the midpoint of BC, the half-turn sends B to C and keeps P fixed, so side PB lands exactly on PC. M goes to a point M′ on line MP extended, with
Now the moved piece \(\triangle M'PC\) and the piece \(\triangle MPC\) share the side PC, and M, P, M′ are on one line (the median AP extended). So the two pieces together make one triangle MCM′ — the angles at P form a straight angle, so there is no bend at P.
Its sides, using the Centroid Theorem (each median is cut 2 : 1 at M):
Do the same about Q (pieces MQC, MQA) and about R (pieces MRA, MRB). Each time you get a triangle whose sides are
So the 6 pieces make 3 congruent triangles (SSS), each with sides two-thirds of the three medians.
*(ii) Let \(a = BC\), \(b = CA\), \(c = AB\) and \(m_a = AP\), \(m_b = BQ\), \(m_c = CR\). Apollonius' relation (it follows from the Pythagoras theorem, by dropping a perpendicular from A to BC) gives
Apply (i) to an assembled triangle, whose sides are
The new pieces have sides two-thirds of its medians. Its median to side x has
Substituting the formula for each median and simplifying, \(2m_b^2 + 2m_c^2 - m_a^2 = \tfrac{9}{4}a^2\), so
Each median of the assembled triangle is half a side of ABC, and two-thirds of that is
The same holds for the other two medians, so each of the 9 final triangles has sides
(iii) Yes. Cut along median AP: pieces \(\triangle APC\) and \(\triangle APB\). Keep \(\triangle APC\). Let N be the midpoint of AB and cut \(\triangle APB\) along its median PN. Turn piece \(\triangle PNB\) through \(180^\circ\) about N: B lands on A and P lands on P′ with N the midpoint of PP′. Pieces PNA and P′NA make triangle PAP′, with sides
So \(\triangle PAP' \cong \triangle APC\) (SSS): two congruent triangles.
A more general midpoint theorem and its converse. In a quadrilateral ABCD, suppose \(AB \parallel DC\). Recall that such ABCD is called a trapezium. Let E be the midpoint of AD. A line drawn through E intersects side BC at F.
(i) Draw the diagonal BD; let EF meet BD at M.
In \(\triangle DAB\): E is the midpoint of AD and \(EM \parallel AB\). By Theorem 7, M is the midpoint of BD and
In \(\triangle BDC\): M is the midpoint of BD and \(MF \parallel DC\) (because \(EF \parallel AB \parallel DC\)). By Theorem 7, F is the midpoint of BC and
E, M, F are on one line, so
(ii) Direct way. Let M be the midpoint of BD. In \(\triangle DAB\), E and M are midpoints, so \(EM \parallel AB\). In \(\triangle BDC\), M and F are midpoints, so \(MF \parallel DC \parallel AB\). The lines EM and MF both pass through M and are both parallel to AB, so they are the same line. Hence E, M, F are collinear and \(EF \parallel AB\).
Using (i). Draw the line through E parallel to AB; by (i) it meets BC at the midpoint of BC, which is F. There is only one line through E parallel to AB, so EF is that line: \(EF \parallel AB\).
Both are short; the second is simpler once (i) is proved, because it only needs uniqueness of the parallel line — exactly like the second proof of Theorem 7.
The diagonals AC and BD of a parallelogram ABCD intersect at O. A line through O meets AB and CD at points P and Q respectively. Show that O is the midpoint of PQ. (Multiple proofs are possible. Which is the simplest?)
Compare \(\triangle OAP\) and \(\triangle OCQ\):
So \(\triangle OAP \cong \triangle OCQ\) (ASA), which gives \(OP = OQ\): O is the midpoint of PQ.
Another proof: a half-turn about O swaps A with C and B with D (diagonals bisect each other), so it maps line AB onto line CD. It maps the line PQ (through O) onto itself. So it sends P = AB ∩ PQ to CD ∩ PQ = Q, hence OP = OQ. The congruence proof is the simplest to write down.
* ABCD is a trapezium with parallel sides AD = 3 cm and BC = 5 cm. E and F are the midpoints of the non-parallel sides. Find the ratio of the areas of the 4-gons AEFD and EBCF.
E and F are the midpoints of AB and DC. By End-of-Chapter Q7, \(EF \parallel AD \parallel BC\) and
Let the height of the trapezium be h. Since E is the midpoint of AB, the line EF is half way between AD and BC, so each strip has height h/2. Both pieces are trapeziums; keep the common height h/2 unevaluated:
Divide. The common factor below cancels from top and bottom:
The ratio is 7 : 9.
Consider 4 points A, B, C, D in the plane with no three collinear. Answer the following questions. Some answers may require you to consider different cases, depending on how the points are positioned in the plane.
(i) A 4-gon on these points is fixed by the cyclic order in which we go round them. Fix A as the start; the other three can follow in \(3 \times 2 \times 1 = 6\) orders, but going round backwards gives the same 4-gon (ABCD = ADCB), so
The three 4-gons are ABCD, ABDC and ACBD — 3 quadrilaterals in every case.
(ii) There are two cases.
| How the points lie | The 3 quadrilaterals |
|---|---|
| No point inside the triangle of the other three | 1 convex, 2 self-intersecting |
| One point inside the triangle of the other three | 3 non-convex (0 convex, 0 self-intersecting) |
Case 1 (the four points are corners of a convex shape): going round the outside in order gives the one convex quadrilateral; each of the other two orders uses both diagonals of it as sides, and they cross — self-intersecting.
Case 2 (say D inside \(\triangle ABC\)): every order joins D to two of A, B, C, giving a triangle with a dent at D. No sides cross, and none is convex, so all 3 are non-convex.
Suppose P is a point on side AB of \(\triangle ABC\) and the line through P parallel to BC meets AC in point Q. For parts (i) to (iii), assume AP = 1, AQ = \(\sqrt{5}\), and see Fig. 12.39.
Key fact. If AB is cut into equal steps and lines parallel to BC are drawn through the cut points, they cut AC into equal steps too. (Go down one step at a time: each strip is a trapezium, and End-of-Chapter Q7 says the parallel through the midpoint of one slanting side passes through the midpoint of the other. The top piece is a triangle, handled by Theorem 7.)
(i) AB = 1 + 2 = 3, so cut AB into 3 steps of 1: A to P is one step, P to B two more. The dotted line through the second step is parallel to BC.
So AC is cut into 3 equal steps, each equal to \(AQ = \sqrt{5}\). QC is two of them:
*(ii) Use steps of length 1/3: AP = 1 is 3 steps and PB is 1 more step, so AB has 4 steps.
AQ covers 3 equal steps of AC, so one step on AC is \(\sqrt{5} \div 3\), and QC is one step:
*(iii) Again use steps of length 1/3: AP = 1 is 3 steps and PB = 2/3 is 2 steps. On AC, AQ is 3 steps, so one step is \(\sqrt{5} \div 3\) and QC is 2 steps:
*(iv) Suppose the ratio is rational, with m and n whole numbers:
Then AP and PB are whole multiples of the same length u:
Cut AB into \(m + n\) steps of u and draw parallels to BC through the cut points. P is a cut point, so PQ is one of these parallels. By the key fact they cut AC into \(m + n\) equal steps, say of length v, and Q is the m-th cut:
Check (i)–(iii): since \(AQ = \sqrt{5}\) and \(AP = 1\), the equal ratios give \(QC = \sqrt{5} \times PB\):
(i) DM \(\parallel\) BN. MB and DN are halves of the equal and parallel sides AB and DC, so
By Theorem 5, MBND is a parallelogram, so \(DM \parallel BN\). Let DM meet AC at X and BN meet AC at Y.
In \(\triangle ABY\): M is the midpoint of AB and \(MX \parallel BY\). By Theorem 7, X is the midpoint of AY:
In \(\triangle CDX\): N is the midpoint of CD and \(NY \parallel DX\). By Theorem 7, Y is the midpoint of CX:
So \(AX = XY = YC\): X and Y trisect AC.
(ii) Three ways to trisect a segment PQ:
1. Parallelogram (from (i)). Take any point B not on line PQ and complete the parallelogram PBQD with PQ as a diagonal (D is where the line through P parallel to BQ meets the line through Q parallel to PB). Mark the midpoints M of PB and N of QD. The lines DM and BN cut PQ into three equal parts.
2. Equal steps and parallels (Exercise 11). Draw any ray from P and mark three equal steps P₁, P₂, P₃ on it with a compass. Join P₃Q. Through P₁ and P₂ draw lines parallel to P₃Q. They cut PQ into three equal parts (AP/PB = AQ/QC with ratios 1 : 2 and 2 : 1).
3. Centroid. Take any point B not on line PQ and let C be the point with Q the midpoint of BC (extend BQ by its own length). Then PQ is a median of \(\triangle PBC\). Draw a second median, from B to the midpoint of PC. It meets PQ at the centroid G, and by the Centroid Theorem
So G is one trisection point; the midpoint of PG is the other.
Is the Midpoint Theorem for Quadrilaterals (Theorem 9) true when the quadrilateral is non-convex? How about when it is self-intersecting? Experiment and check. The pictures may look strange, but the theorem appears to be true.
(i) Yes, it is true in both cases. The proof in the text never used convexity. It only applied the Midpoint Theorem to the four triangles ABC, ADC, BCD and BAD:
Each of these triangles exists whatever the shape of ABCD (no three vertices are collinear), so PQRS has both pairs of opposite sides parallel. Only the picture changes: for a self-intersecting ABCD the parallelogram may lie partly outside it.
(ii) PQ is parallel to AC and PS is parallel to BD. If the diagonals AC and BD are parallel to each other, then PQ and PS are parallel lines through the same point P, so P, Q, R, S all lie on one line and the “parallelogram” is flattened into a segment. (It can never shrink to a point, because \(PQ = \tfrac{1}{2}AC\) is not zero.)
This happens exactly when AC \(\parallel\) BD. Then A, C, D, B are the corners of a trapezium in that order, so ABCD (which uses AC and BD as its “diagonals”) is a self-intersecting 4-gon. For an ordinary (convex or non-convex) quadrilateral the diagonals are never parallel, so this exception never arises.
*(iii) Yes. Each of the four triangles ABC, ADC, BCD, BAD is still a flat triangle (three points always lie in a plane), so the Midpoint Theorem still gives \(PQ \parallel AC \parallel SR\) and \(PQ = SR = \tfrac{1}{2}AC\). Two parallel lines PQ and SR lie in one plane, so P, Q, R, S are coplanar, and a quadrilateral with a pair of equal and parallel opposite sides is a parallelogram (Theorem 5). So the Varignon parallelogram of a non-planar 4-gon is a genuine flat parallelogram.
Let P, Q, R, S be four points on sides AB, BC, CD and DA respectively of a quadrilateral ABCD. Suppose PQRS is a parallelogram. Must P, Q, R and S be midpoints of the respective sides? This can be considered a possible converse question to Theorem 9.
No. Counterexample: let ABCD be a square of side 3 and take each point one-third of the way along its side (going round in the same direction):
The four corner triangles \(\triangle SAP\), \(\triangle PBQ\), \(\triangle QCR\), \(\triangle RDS\) are congruent (SAS: legs 1 and 2 with a right angle between), so
PQRS has all sides equal, so it is a parallelogram (even a square, tilted) — yet none of P, Q, R, S is a midpoint. The same works with any fraction in place of one-third, and in any parallelogram ABCD.
(i) P, Q are the midpoints of AB, AC. To get \(\triangle APQ \cong \triangle CSQ\) by SAS we need QS = QP, so extend PQ by its own length: choose S with \(QS = PQ\). Then
so \(\triangle APQ \cong \triangle CSQ\) (SAS). Hence
The equal alternate angles (transversal AC) give \(CS \parallel AB\), i.e. \(CS \parallel PB\). So PBCS has a pair of equal and parallel opposite sides, and is a parallelogram (Theorem 5). Therefore
Since \(PQ = \tfrac{1}{2}PS\):
(ii) Now P is the midpoint of AB and the line through P parallel to BC meets AC at Q. Let S be the point where line PQ meets the line through C parallel to AB. Then PBCS has both pairs of opposite sides parallel, so it is a parallelogram and
In \(\triangle APQ\) and \(\triangle CSQ\): \(AP = CS\), \(\angle AQP = \angle CQS\) (vertically opposite) and \(\angle QAP = \angle QCS\) (alternate angles, \(CS \parallel AB\)). So \(\triangle APQ \cong \triangle CSQ\) (AAS), giving
So Q is the midpoint of AC, which is Theorem 7.
* Review all the properties of a rhombus/rectangle/square that you proved in Grade 8. Formulate a converse of each. Decide if the converse is true. There are many possibilities here!
Below, “4-gon” means any quadrilateral, so each converse starts from nothing but the property.
| Converse | True? |
|---|---|
| A 4-gon with all four sides equal is a rhombus. | Yes. Opposite sides equal ⇒ parallelogram (Theorem 2), with equal sides. |
| A 4-gon whose diagonals are perpendicular is a rhombus. | No. A kite: A(0, 3), B(−2, 0), C(0, −1), D(2, 0). |
| A 4-gon whose diagonals are perpendicular bisectors of each other is a rhombus. | Yes. Exercise Set 12.2, Q3 (ii). |
| A 4-gon whose diagonals bisect its angles is a rhombus. | Yes. Exercise Set 12.2, Q3 (i). |
| A 4-gon with all four angles equal is a rectangle. | Yes. Each angle is \(360^\circ / 4 = 90^\circ\). |
| A 4-gon with equal diagonals is a rectangle. | No. Isosceles trapezium A(0, 0), B(6, 0), C(4, 3), D(2, 3). |
| A 4-gon whose diagonals are equal and bisect each other is a rectangle. | Yes. Exercise Set 12.2, Q3 (iii). |
| A 4-gon with equal and perpendicular diagonals is a square. | No. The same kite: both diagonals are 4 and perpendicular, sides unequal. |
| A 4-gon whose diagonals are equal, perpendicular and bisect each other is a square. | Yes. Rhombus (Q3 (ii)) with equal diagonals is also a rectangle. |
| A rhombus with one right angle is a square. | Yes. It is a parallelogram with a right angle, so all angles are right angles. |
| A rectangle with two adjacent sides equal is a square. | Yes. Opposite sides are equal too, so all four sides are equal. |
* Show that the sum of angles of a non-planar quadrilateral is always less than \(360^\circ\). Can you find a non-planar quadrilateral ABCD for which \(\angle A + \angle B + \angle C + \angle D = 2^\circ\)? (Hint: Think of a diagonal, say AC, as a hinge around which triangles ABC and ADC can rotate.) What happens to each angle of ABCD as you do this rotation?
Use the diagonal AC as a hinge. The triangles ABC and ADC are flat, and their angles add up to \(180^\circ\) each:
At A, the three rays AB, AC, AD do not lie in one plane (ABCD is non-planar). For three such rays the angle between two of them is less than the sum of the other two (the shortest path on a sphere is the direct one), so
Add everything:
What happens during the rotation. \(\angle B\) and \(\angle D\) never change (they belong to the rigid triangles). When the hinge is fully open (ABCD flat and convex) the sum is \(360^\circ\). As we fold \(\triangle ADC\) towards \(\triangle ABC\), \(\angle A\) and \(\angle C\) shrink steadily; when the two triangles lie on top of each other they become \(|\angle BAC - \angle DAC|\) and \(|\angle BCA - \angle DCA|\).
A sum of \(2^\circ\). Take two congruent tall thin triangles ABC and ADC on the base AC = 1, with apex angles \(\angle B = \angle D\) a little under \(1^\circ\) (for example apex 100 units above the middle of AC gives about \(0.57^\circ\)). Folded flat on top of each other (D on B), \(\angle A\) and \(\angle C\) become \(0^\circ\) and the sum is \(\angle B + \angle D\), about \(1.15^\circ\). Fully open, it is \(360^\circ\). As the hinge opens the sum changes continuously from about \(1.15^\circ\) to \(360^\circ\), so at some opening it is exactly \(2^\circ\), and at that moment ABCD is non-planar.
* Let us see a third method to tile the plane using a 4-gon. Focus on only two coloured copies of SOME sharing a vertex. What do you see? It appears that each 4-gon is just a shifted copy of the other. Let us see exactly how. Draw two copies of SOME as shown in Fig. 12.42 so that the diagonals EO and E′O′ are collinear with O = E′. Now place a cutout of one copy on top of SOME with diagonal EO drawn on it. Slide this cutout so that segment EO moves along EO′ until EO matches E′O′. Verify that your cutout exactly matches S′O′M′E′.
(i) Sliding the cutout along EO by the length EO moves every point the same distance in the same direction. So each vertex X of SOME goes to X′ with
(for E this gives E′ = O). Now look at the four quadrilaterals
In each one, the two sides XX′ and YY′ are equal and parallel (both equal and parallel to EO). By Theorem 5 each is a parallelogram, so its other pair of sides are equal and parallel too:
All four sides of S′O′M′E′ are equal and parallel to those of SOME, and the same parallelogram argument on the diagonal (E E′ O′ O is a straight strip with \(E'O' = EO\)) gives the diagonal too. So \(\triangle S'O'E' \cong \triangle SOE\) and \(\triangle M'O'E' \cong \triangle MOE\) (SSS), and the slid cutout exactly covers S′O′M′E′.
(ii) Slide along the diagonal EO by EO forwards and backwards, and along the other diagonal SM by SM forwards and backwards: that places 4 copies around the starting one, each touching it only at a vertex (O of one copy is E of the next). Repeating from these copies fills the 3 by 3 grid. Each copy is the starting one shifted by a whole number of steps EO and SM. The blank 4-gon between four neighbouring copies has sides that are shifted copies of the four sides of SOME, taken in the opposite order — it is SOME turned through \(180^\circ\), so it is congruent to SOME.
* Is there a 4-gon with given side lengths?
(i) Draw BC = c. Draw a circle of radius b about C and a circle of radius a about B. A triangle exists exactly when these circles meet off the line BC. They meet when the distance between centres is less than the sum of radii and more than their difference: \(b - a < c < a + b\). With \(a \le b \le c\) the left part is automatic, so the test is \(a + b > c\).
(ii) Rule: in a 4-gon each side is shorter than the other three put together (the straight side is the shortest path between its ends; the other three sides form a longer path). Let the fourth side be d.
So \(4 < d < 18\).
• d = 100: No (100 > 18).
• d = 10: Yes. Use a diagonal of length 5: triangles with sides 2, 5, 5 and 11, 10, 5 both exist, so put them on opposite sides of the diagonal.
• d = 1: No (2 + 5 + 1 = 8 is less than 11).
The possible lengths are all d with \(4 < d < 18\).
(iii) Arrange them as \(a \le b \le c \le d\). A 4-gon exists exactly when \(d < a + b + c\) (the longest side is shorter than the other three together).
Necessity is the rule in (ii). For sufficiency, split the 4-gon by a diagonal of length x into triangles (a, d, x) and (b, c, x). By (i) both triangles exist when
Such an x exists because \(d - a < b + c\) (that is the condition \(d < a + b + c\)) and the other comparisons are automatic (\(c - b < a + d\), \(d - a < a + d\), \(c - b < b + c\)). Choose such an x, build both triangles on it on opposite sides, and you have a 4-gon with sides a, d, c, b.
* Counting diagonals of a polygon.
(i) A diagonal of an n-gon is a segment joining two vertices that are not adjacent (not the two ends of a side).
| n | diagonals | added |
|---|---|---|
| 3 | 0 | — |
| 4 | 2 | 2 |
| 5 | 5 | 3 |
| 6 | 9 | 4 |
| 7 | 14 | 5 |
| 8 | 20 | 6 |
The jumps are 2, 3, 4, 5, 6, so the guesses for n = 7 and 8 are 14 and 20; counting vertex by vertex confirms them (below).
(ii) From each vertex, diagonals go to every vertex except itself and its two neighbours: \(n - 3\) of them. n vertices give \(n(n - 3)\), but each diagonal has two ends and was counted twice:
Check with n = 7 and n = 8:
Adding one side. Insert a new vertex between two neighbours X and Y. The old side XY becomes a diagonal (1 more), and the new vertex sends diagonals to all old vertices except X and Y (\(n - 2\) more). Total added:
Starting from 0 diagonals for n = 3 and adding 2, 3, ..., \(n - 2\) gives
which proves the formula for every n.
* Sum of angles of a polygon. What is the sum of angles of a (planar non-self-intersecting) n-gon? We know that the answer is \(180^\circ\) for n = 3 and \(360^\circ\) for n = 4. Find the next few values. Then find a formula in terms of n and prove it.
Cut the n-gon into triangles using diagonals that lie inside it. Every polygon can be cut this way into exactly \(n - 2\) triangles: for a convex polygon, draw all diagonals from one vertex; for a non-convex one, there is always some diagonal lying inside, and cutting along it splits the polygon into two smaller ones, so we can keep going.
The angles of the triangles together make up exactly the angles of the polygon (the cuts run between vertices), so
| n | angle sum |
|---|---|
| 3 | \(180^\circ\) |
| 4 | \(360^\circ\) |
| 5 | \(540^\circ\) |
| 6 | \(720^\circ\) |
| 7 | \(900^\circ\) |
| 8 | \(1080^\circ\) |
Why exactly \(n - 2\) triangles: every cut along a diagonal splits a k-gon into two pieces whose vertex counts add to \(k + 2\) (the two ends of the diagonal are shared). If each piece with m vertices gives \(m - 2\) triangles, then together they give \((m_1 - 2) + (m_2 - 2) = (k + 2) - 4 = k - 2\). Starting from triangles (\(3 - 2 = 1\)) this holds for every n.
* Multiple converses to a theorem. Let us see how multiple statements can be considered converses to the Midpoint Theorem and how Theorem 7 is one of them. To formulate a converse we should express the original statement in “If ... then ...” form. For a complex statement, there may be multiple ways to do that. The Midpoint Theorem starts with \(\triangle ABC\) and points P and Q on sides AB and AC respectively. The theorem has two assumptions and two conclusions, which we have named for further discussion. Assumptions: (P MID) P is the midpoint of AB, and (Q MID) Q is the midpoint of AC. Conclusions: (PRLL) \(PQ \parallel BC\), and (HALF) PQ = BC/2. Let us use these four named conditions to discuss various possible statements.
(i) Statement: In \(\triangle ABC\), P is on AB and Q is on AC. If \(PQ \parallel BC\) and \(PQ = \tfrac{1}{2}BC\), then P and Q are the midpoints of AB and AC.
Proof. Through Q draw the line parallel to AB, meeting BC at R. PQRB has both pairs of opposite sides parallel, so it is a parallelogram, and
So R is the midpoint of BC and \(RC = \tfrac{1}{2}BC = PQ\). Now compare \(\triangle APQ\) and \(\triangle QRC\):
So \(\triangle APQ \cong \triangle QRC\) (ASA). Hence \(AQ = QC\) and \(AP = QR = PB\): P and Q are midpoints.
(ii) Converse of the second sentence, keeping the first: “Suppose P is the midpoint of side AB of \(\triangle ABC\) and Q is a point on side AC. If \(PQ \parallel BC\) then Q is the midpoint of AC.” This is exactly Theorem 7, and the proof of Theorem 7 also gave \(PQ = \tfrac{1}{2}BC\). In named conditions:
(iii) Choosing two of the four conditions as assumptions gives 6 choices:
| Assume | Status |
|---|---|
| (P MID), (Q MID) | the Midpoint Theorem |
| (PRLL), (HALF) | part (i) — true |
| (P MID), (PRLL) | Theorem 7 |
| (Q MID), (PRLL) | Theorem 7 with B, C and P, Q swapped |
| (Q MID), (HALF) | the same as (P MID), (HALF) with the roles swapped |
| (P MID), (HALF) | the one left to examine |
In words: “In \(\triangle ABC\), P is the midpoint of AB and Q is a point on AC with \(PQ = \tfrac{1}{2}BC\). Must Q be the midpoint of AC, and must PQ be parallel to BC?”
Answer: not necessarily. The midpoint N of AC always works (PN = BC/2), but the circle with centre P and radius \(\tfrac{1}{2}BC\) can meet side AC at a second point. Example: A(0, 0), B(3, 3), C(4, 0); P = (1.5, 1.5) and Q = (1, 0).
So \(PQ = \tfrac{1}{2}BC\), Q is on AC, yet Q is not the midpoint (2, 0) of AC and PQ is not parallel to BC. So (P MID) and (HALF) do not force (Q MID) or (PRLL).
* Validity of tiling methods. Show using reasoning that each of the three tiling methods we saw produces a tiling of the plane using the given 4-gon. You have to prove that when we place new copies using any of the three procedures, the entire plane is covered with no gaps and no overlaps. First, think carefully about what it means to prove this. Then find a proof for each procedure.
What must be proved. (a) Every point of the plane lies in some copy (no gaps), and (b) no point lies inside two copies (no overlaps). A neat way is to show that the copies fill, without overlap, a parallelogram that itself tiles the plane by shifting — because parallelograms tile the plane (introduction).
Method 3 (sliding, EoC Q18). Let \(u = \overrightarrow{EO}\) and \(v = \overrightarrow{SM}\). All coloured copies are SOME shifted by \(ju + kv\) for whole numbers j, k, and all blank 4-gons are SOME turned through \(180^\circ\) and shifted the same way. The parallelogram with sides u and v has area equal to twice the area of SOME, and it is made up of exactly one coloured piece and one blank piece. Since the shifts of that parallelogram tile the plane, so do the 4-gons.
Method 1 (half-turns about midpoints). Two half-turns, about points \(M_1\) then \(M_2\), add up to a shift through \(2\overrightarrow{M_1M_2}\). So turning about the midpoints of two different sides of SOME produces copies that are shifted versions of the starting copy or of its half-turned copy — the same family of copies as Method 3. That is why “each new copy can be obtained by rotating any one of its neighbours, and both ways give the same result” and why the copies fit perfectly. Around each vertex the angles 1, 2, 3, 4 appear once each, adding to \(360^\circ\).
Method 2 (Varignon grid). By Exercise Set 12.3 Q5, SOME is fixed once its Varignon parallelogram and one vertex are known. The grid of Varignon parallelograms is a parallelogram tiling. Each copy of SOME sits on one grid parallelogram plus four corner triangles that stick out; the corner triangles from four neighbouring copies around a gap form exactly the half-turned copy of SOME in that gap. Since the grid covers the plane without gaps or overlaps and every grid cell is shared out exactly once, so do the copies of SOME.
* What fraction of the square is shaded?
Take the square ABCD with side 2, so its area is 4. K, L, M, N are the midpoints of AB, BC, CD, DA. The four lines DK, MB, NC and AL enclose the shaded square.
Cut and rearrange. Each line, such as DK, cuts off a right triangle (like \(\triangle DAK\)) with legs 2 and 1. Around the shaded square there are four such right triangles with a small corner triangle cut off each; the small corner triangle fits exactly into the notch of the next piece. So the square is made up of the shaded square and four pieces that rearrange into four more copies of the shaded square. Hence the shaded part is one of five equal squares.
By calculation. The side DK of the big right triangle is
The distance between the parallel lines DK and MB is the side s of the shaded square. The parallelogram KBMD has area
and also base DK times this distance:
The shaded part is \(\tfrac{1}{5}\) of the square.
Your orientation request has been received.
Saraswati Vidyamandir will contact you soon.
