Triangles

The whole exercise turns on one distinction, so it is worth stating before filling anything in: congruent means same shape and same size; similar means same shape, size not necessarily the same. Every congruent pair is similar, but not the other way round.

(i) Two circles can have different radii, so they need not be the same size — but a circle has only one possible shape, so any two circles must have the same shape. Therefore all circles are similar. (They are congruent only in the special case of equal radii.)

(ii) Same reasoning: every square has four right angles and four equal sides, so all squares have the same shape, but a 2 cm square and a 5 cm square are different sizes. All squares are similar.

(iii) An equilateral triangle always has all three angles \(60^{\circ}\), so any two equilateral triangles have equal corresponding angles and sides in the same ratio. An isosceles triangle does not pin the shape down — one can have apex angle \(20^{\circ}\) and another \(120^{\circ}\). So all equilateral triangles are similar.

(iv) This is the formal definition of similarity for polygons: corresponding angles equal, corresponding sides proportional.

Answers: (i) similar  ·  (ii) similar  ·  (iii) equilateral  ·  (iv) (a) equal, (b) proportional.

To answer this properly, each example has to be tested against the definition — corresponding angles equal and corresponding sides in the same ratio — not just eyeballed.

(i) Two pairs of similar figures

1. Any two equilateral triangles, say of sides 3 cm and 5 cm. Every angle in each is \(60^{\circ}\), so corresponding angles are equal; and the three side ratios are \(\dfrac{3}{5},\ \dfrac{3}{5},\ \dfrac{3}{5}\) — all the same. Both conditions hold, so they are similar.

2. Any two squares, say of sides 2 cm and 7 cm. Every angle in each is \(90^{\circ}\), and all four side ratios equal \(\dfrac{2}{7}\). Similar.

(Two circles of different radii work equally well, as does a photograph and its enlargement — NCERT’s own Taj Mahal example on page 75.)

(ii) Two pairs of non-similar figures

1. A square and a rectangle, say a \(3\ \text{cm}\times3\ \text{cm}\) square and a \(3\ \text{cm}\times5\ \text{cm}\) rectangle. All corresponding angles are \(90^{\circ}\), so condition (i) holds — but the side ratios are \(\dfrac{3}{3}=1\) and \(\dfrac{3}{5}\), which are different, so condition (ii) fails. Equal angles alone are not enough.

2. A square and a rhombus whose sides are all the same length, say 4 cm each, but whose angles are \(90^{\circ}\) in the square and \(60^{\circ}\) and \(120^{\circ}\) in the rhombus. Now every side ratio is \(\dfrac{4}{4}=1\), so condition (ii) holds — but the angles differ, so condition (i) fails. Proportional sides alone are not enough either.

Those last two examples are exactly NCERT’s Fig. 6.6 and Fig. 6.7, and together they make the chapter’s point: both conditions are needed.

Step 1 — check the sides. Every side of rhombus PQRS is 1.5 cm and every side of square ABCD is 3 cm, so taking the sides in corresponding order:

All four ratios are equal, so the corresponding sides are proportional. Condition (ii) of the definition is satisfied.

Step 2 — check the angles. This is where it breaks down. Every angle of the square ABCD is \(90^{\circ}\). The figure PQRS is a rhombus that is not a square: its angles come in two unequal pairs (an acute pair and an obtuse pair), so they are not \(90^{\circ}\). Therefore

and the corresponding angles are not equal. Condition (i) fails.

Conclusion. Similarity needs both conditions, and only one of them holds here. So the two quadrilaterals are not similar.

This is the same trap as NCERT’s Fig. 6.7 on page 78: having all sides in the same ratio is not enough on its own — a square and a rhombus of equal side lengths have every side ratio equal to 1 and are still not similar.

Both parts are direct applications of the Basic Proportionality Theorem (Theorem 6.1): since DE ∥ BC and DE cuts AB at D and AC at E,

(i) Here \(AD=1.5\) cm, \(DB=3\) cm and \(AE=1\) cm; we need EC. Substituting:

Cross-multiplying:

So EC = 2 cm.

(ii) Here \(DB=7.2\) cm, \(AE=1.8\) cm and \(EC=5.4\) cm; we need AD. Substituting:

The right-hand side simplifies first — always worth doing before multiplying out:

So:

So AD = 2.4 cm.

The tool for every part is Theorem 6.2, the converse of the Basic Proportionality Theorem: EF is parallel to QR if and only if EF divides PQ and PR in the same ratio, i.e. if and only if \(\dfrac{PE}{EQ}=\dfrac{PF}{FR}\). So in each case we compute the two ratios and compare them.

(i) \(PE=3.9,\ EQ=3,\ PF=3.6,\ FR=2.4\).

Since \(1.3\neq1.5\), the two ratios are different, so EF is NOT parallel to QR.

(ii) \(PE=4,\ EQ=4.5,\ PF=8,\ FR=9\).

The two ratios are equal, so by Theorem 6.2, EF ∥ QR.

(iii) Careful here — this part gives the WHOLE sides PQ and PR, not the second pieces EQ and FR, so those must be worked out first:

Now compare the ratios:

They are equal, so EF ∥ QR.

Note: in part (iii) one could equally compare \(\dfrac{PE}{PQ}\) with \(\dfrac{PF}{PR}\) — both come to \(\dfrac{9}{64}\) — which is the form proved in Example 1 and avoids the subtraction altogether.

Step 1 — use the first parallel, in △ABC. Here LM ∥ CB, with M on AB and L on AC. This is exactly the situation of Example 1, whose result (proved there from Theorem 6.1) is that the line cuts off proportional parts of the whole sides:

Call this equation (1).

Step 2 — use the second parallel, in △ACD. Now LN ∥ CD, with N on AD and L on AC. By exactly the same argument in this second triangle:

Call this equation (2).

Step 3 — combine. The right-hand sides of (1) and (2) are the same quantity, \(\dfrac{AL}{AC}\). Two things equal to the same thing are equal to each other:

Hence proved. Notice the shape of the argument: the segment AL on the shared line AC is the common link between two otherwise unrelated triangles — the same bridging idea used with the diagonal in Example 2.

Step 1 — apply Theorem 6.1 in △ABC. In this triangle the line DE is parallel to the side AC, and it cuts BA at D and BC at E. So the two sides it cuts are divided in the same ratio:

Call this equation (1).

Step 2 — apply Theorem 6.1 again, but in △ABE. This is the step that is easy to miss: DF is parallel to AE, and inside triangle ABE the line DF cuts BA at D and BE at F. So:

Call this equation (2).

Step 3 — combine. Both (1) and (2) have the same left-hand side, \(\dfrac{BD}{DA}\). Therefore their right-hand sides are equal:

Hence proved. The ratio \(\dfrac{BD}{DA}\) on side AB is doing all the work — it is the one quantity that both parallel lines have in common, because both of them cut AB at the same point D.

Step 1 — apply Theorem 6.1 in △POQ. Inside this triangle, DE is parallel to the side OQ and cuts PO at D and PQ at E:

Call this equation (1).

Step 2 — apply Theorem 6.1 in △POR. Here DF is parallel to the side OR and cuts PO at D and PR at F:

Call this equation (2).

Step 3 — combine (1) and (2). Both have the same left-hand side, so:

Step 4 — run the converse, in △PQR. We have just shown that the line EF divides the two sides PQ and PR of triangle PQR in the same ratio. By Theorem 6.2 (the converse of the Basic Proportionality Theorem), such a line must be parallel to the third side:

Hence shown. The pattern here is worth naming, because Q6 repeats it exactly: use Theorem 6.1 twice to produce a common ratio, then use Theorem 6.2 once to convert that ratio back into a parallel line.

Step 1 — apply Theorem 6.1 in △OPQ. The line AB is parallel to the side PQ and cuts OP at A and OQ at B:

Call this equation (1).

Step 2 — apply Theorem 6.1 in △OPR. The line AC is parallel to the side PR and cuts OP at A and OR at C:

Call this equation (2).

Step 3 — combine. Both left-hand sides are the same ratio \(\dfrac{OA}{AP}\) on the shared segment OP, so:

Step 4 — apply the converse in △OQR. The line BC now divides the two sides OQ and OR of triangle OQR in the same ratio, so by Theorem 6.2:

Hence shown.

Setting it up. Take △ABC. Let D be the mid-point of AB, and let the line through D parallel to BC meet AC at E. We have to prove that E is the mid-point of AC, that is, \(AE=EC\).

Step 1 — write down what “mid-point” gives us. Since D is the mid-point of AB:

Dividing both sides by DB turns this into the ratio form we will need:

Call this equation (1).

Step 2 — apply Theorem 6.1. The line DE is parallel to BC and cuts the other two sides AB and AC at the distinct points D and E. So:

Call this equation (2).

Step 3 — substitute (1) into (2).

Step 4 — read off the conclusion. A ratio equal to 1 means the two quantities are equal:

So E is the mid-point of AC, i.e. the line bisects the third side. Hence proved.

This is the Mid-point Theorem from Class IX, now obtained as a one-line special case of Theorem 6.1 — the special case where the dividing ratio happens to be \(1:1\).

Setting it up. Take △ABC. Let D be the mid-point of AB and E the mid-point of AC. We have to prove that DE ∥ BC.

Notice this is the reverse of Q7: there we were given the parallel and had to prove the bisection; here we are given both bisections and must prove the parallel. So the tool changes from Theorem 6.1 to its converse, Theorem 6.2.

Step 1 — write both mid-point facts as ratios. Since D is the mid-point of AB:

Since E is the mid-point of AC:

Step 2 — compare the two ratios. Both are equal to 1, so they are equal to each other:

Step 3 — apply Theorem 6.2. The line DE divides the two sides AB and AC of the triangle in the same ratio. By the converse of the Basic Proportionality Theorem, it must therefore be parallel to the third side:

Hence proved.

Step 1 — draw the helping line. Nothing in the figure is yet a triangle with a line parallel to one of its sides, so we make one. Through O draw a line EO parallel to AB, with E on AD.

Step 2 — note the second parallel. We are given AB ∥ DC, and we have just drawn EO ∥ AB. Lines parallel to the same line are parallel to each other, so:

Step 3 — apply Theorem 6.1 in △ADC. Here EO ∥ DC, cutting AD at E and AC at O:

Call this equation (1).

Step 4 — apply Theorem 6.1 in △ABD. Here EO ∥ AB, cutting DA at E and DB at O:

Inverting both sides so it lines up with equation (1):

Call this equation (2).

Step 5 — combine (1) and (2). Both equal \(\dfrac{AE}{ED}\), so:

Step 6 — rearrange into the form asked for. Cross-multiplying gives \(AO\times OD=BO\times OC\), and dividing both sides by \(BO\times OD\):

Hence shown.

This is the converse of Q9, so we expect to finish with Theorem 6.2 rather than Theorem 6.1. Recall that a trapezium is a quadrilateral with one pair of opposite sides parallel — so proving AB ∥ DC is the whole job.

Step 1 — draw the helping line. Through O draw a line OE parallel to AB, meeting AD at E.

Step 2 — apply Theorem 6.1 in △ABD. Here OE ∥ AB, cutting DA at E and DB at O:

Call this equation (1).

Step 3 — rewrite what was given. We are told \(\dfrac{AO}{BO}=\dfrac{CO}{DO}\). Cross-multiplying, \(AO\times DO=BO\times CO\), and dividing both sides by \(CO\times DO\) rearranges it into:

Inverting both sides:

Call this equation (2).

Step 4 — combine (1) and (2). Both right-hand sides are \(\dfrac{DO}{OB}\), so:

Step 5 — apply Theorem 6.2 in △ADC. The equation above says that the line EO divides the two sides DA and DC… more precisely, it divides AD at E and AC at O in the same ratio. By the converse of the Basic Proportionality Theorem:

Step 6 — finish. But EO was drawn parallel to AB in Step 1. Since EO is parallel to both AB and DC, those two lines must be parallel to each other:

So ABCD has a pair of parallel opposite sides, i.e. ABCD is a trapezium. Hence shown.

Each part is decided by testing the three criteria in turn: AAA/AA (all angles equal), SSS (all three side ratios equal) and SAS (two side ratios equal with the angle between them equal). The correspondence of vertices must then be written in the right order.

(i)

In △ABC the angles are \(60^{\circ},80^{\circ},40^{\circ}\) at A, B, C; in △PQR they are \(60^{\circ},80^{\circ},40^{\circ}\) at P, Q, R. Matching equal angles:

All three pairs of corresponding angles are equal. Similar, by the AAA criterion: \(\triangle ABC\sim\triangle PQR\).

(ii)

All six sides are given, so test SSS. Pair each side of the small triangle with the side of the large one that keeps the ratios equal:

All three ratios equal \(\dfrac{1}{2}\). Similar, by the SSS criterion: \(\triangle ABC\sim\triangle QRP\). Note the order — A goes with Q, B with R and C with P, which is what the ratios above dictate.

(iii)

Again all six sides are known, so test SSS. Order each triangle’s sides from smallest to largest so the best possible pairing is used: for △LMP they are \(2,\ 2.7,\ 3\); for △DEF they are \(4,\ 5,\ 6\). The ratios are then:

The middle ratio, \(0.54\), does not match the other two. Since even the most favourable pairing fails, no pairing can work. Not similar.

(iv)

Here two sides and the angle between them are given in each triangle, which points straight at SAS:

and the included angles are equal:

∠M lies between MN and ML, and ∠Q between QP and QR — exactly the four sides used above. Similar, by the SAS criterion: \(\triangle MNL\sim\triangle QPR\).

(v)

This one looks like part (iv) but is not, and the difference is the whole point of the question. In △ABC we are given \(AB=2.5\), \(BC=3\) and \(\angle A=80^{\circ}\). The angle ∠A lies between AB and AC — but AC is not given. So the known angle is not the angle included between the two known sides, and SAS cannot be applied.

(In △DEF the given angle ∠F is included between the given sides DF and EF, but a criterion has to hold in both triangles.) With only one angle known in each and no matching pair of ratios around it, none of AAA, SSS or SAS applies. Not similar — more precisely, the given data is not enough to conclude similarity.

(vi)

Only angles are given, so find the missing third angle in each triangle using the angle sum property.

In △DEF, \(\angle D=70^{\circ}\) and \(\angle E=80^{\circ}\):

In △PQR, \(\angle Q=80^{\circ}\) and \(\angle R=30^{\circ}\):

Now match them up: \(\angle D=\angle P=70^{\circ}\), \(\angle E=\angle Q=80^{\circ}\) and \(\angle F=\angle R=30^{\circ}\). Similar, by the AA criterion: \(\triangle DEF\sim\triangle PQR\). (Two matching angles were enough; the third was guaranteed to follow.)

Step 1 — find ∠DOC using the straight line. The points D, O and B lie on one straight line, so ∠DOC and ∠BOC are angles on a straight line and add up to \(180^{\circ}\):

Step 2 — find ∠DCO using the angle sum of △DOC. We now know two of its three angles: \(\angle CDO=70^{\circ}\) (given) and \(\angle DOC=55^{\circ}\) (just found).

Step 3 — find ∠OAB using the similarity. We are given △ODC ~ △OBA. Reading the correspondence letter by letter: O↔O, D↔B, C↔A. So the angle at C in the first triangle equals the angle at A in the second:

Answers: ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°.

Notice the instruction: this must be done with a similarity criterion, not with the Basic Proportionality Theorem (which is how Exercise 6.2 Q9 handled the same configuration). So we look for two similar triangles containing OA, OC, OB and OD — namely △AOB and △COD.

Step 1 — first pair of equal angles. Since AB ∥ DC and AC is a transversal cutting them, ∠OAB and ∠OCD are alternate interior angles:

Step 2 — second pair of equal angles. Now take BD as the transversal cutting the same parallel lines, giving another pair of alternate interior angles:

Step 3 — apply the AA criterion. Two pairs of corresponding angles are equal, which is enough:

(As a check, the third pair ∠AOB and ∠COD are vertically opposite, so they are equal too — consistent with the similarity.)

Step 4 — write the ratio of corresponding sides. With the correspondence A↔C, O↔O, B↔D:

Hence shown.

Step 1 — turn the angle condition into a side condition. The angles marked ∠1 and ∠2 are ∠PQR and ∠PRQ, the angles at Q and R in △PQR. Since they are equal, the sides opposite them are equal. The side opposite ∠PQR is PR, and the side opposite ∠PRQ is PQ:

Call this equation (1). This step is the key to the whole question — it is what lets the given ratio be rewritten in usable form.

Step 2 — substitute into the given ratio. We are given:

Replacing PR by PQ, using equation (1):

Step 3 — rearrange so the sides of each triangle stay together. Cross-multiplying gives \(QR\times PQ=QS\times QT\), and dividing both sides by \(QT\times QR\):

Call this equation (2). Now read it as a statement about two triangles: PQ and QS are the two sides of △PQS meeting at Q, while QT and QR are the two sides of △TQR meeting at Q.

Step 4 — the included angle. S lies on QR and P lies on QT, so the angle at Q is one and the same angle in both triangles:

Call this equation (3). It is the angle between the very sides paired up in equation (2).

Step 5 — apply SAS. Equations (2) and (3) give two pairs of proportional sides with equal included angles, so by the SAS similarity criterion:

Hence shown.

This is a two-line proof once the right pair of angles is spotted, so the work is all in choosing them.

Step 1 — the given pair. The question hands us one equality directly:

(∠P means ∠RPQ, the angle of △PQR at P.) Call this equation (1).

Step 2 — the free pair. Look at the two triangles RPQ and RTS. Both of them have a vertex at R, and since S lies on PR and T lies on QR, the angle at R is the same angle in both:

Call this equation (2). A shared angle costs nothing to use and is very often the second pair you need.

Step 3 — apply the AA criterion. Two pairs of corresponding angles are equal, which by Theorem 6.3’s AA form is enough for similarity:

Hence shown. Check the correspondence against the two equalities used: R pairs with R from equation (2), and P pairs with T from equation (1), which forces Q to pair with S — exactly the order asked for.

Congruent triangles give equal parts, not merely proportional ones — so the plan is to extract two equal-length facts from the congruence and turn them into a ratio.

Step 1 — read off the corresponding parts. From △ABE ≅ △ACD, matching the letters in order (A↔A, B↔C, E↔D):

Call these equations (1) and (2).

Step 2 — build the ratio. Divide equation (2) by equation (1):

Rearranging so the sides of each triangle stay together — AD and AE belong to △ADE, while AB and AC belong to △ABC:

Call this equation (3).

Step 3 — the included angle. Since D lies on AB and E lies on AC, the angle at A is the same angle in both triangles:

Call this equation (4). It is the angle between the pairs of sides used in equation (3).

Step 4 — apply SAS. By the SAS similarity criterion:

Hence shown.

Every part is an AA argument, and every one of them uses the same two ingredients: a right angle (because AD and CE are altitudes, so \(\angle ADB=\angle ADC=90^{\circ}\) and \(\angle CEA=\angle CEB=90^{\circ}\)) plus one more angle that is either vertically opposite or shared.

(i) △AEP ~ △CDP

The right angles:

The second pair — ∠APE and ∠CPD are vertically opposite angles, since AD and CE cross at P:

Two pairs of equal angles, so by AA:

(ii) △ABD ~ △CBE

The right angles:

The second pair — ∠B is common to both triangles:

By AA:

(iii) △AEP ~ △ADB

The right angles:

The second pair — ∠A is common (∠PAE and ∠BAD are the same angle, since P lies on AD and E lies on AB):

By AA:

(iv) △PDC ~ △BEC

The right angles:

The second pair — ∠C is common (∠DCP and ∠ECB are the same angle, since P lies on CE and D lies on CB):

By AA:

All four shown. The pattern worth carrying away: in a figure full of altitudes, every pair of triangles already shares a right angle, so you only ever need to find one more equal angle — and it will be either a common angle or a vertically opposite one.

Step 1 — use the parallelogram’s opposite angles. In a parallelogram, opposite angles are equal:

Written out in full for the two triangles we care about, that is \(\angle BAE=\angle FCB\). Call this equation (1).

Step 2 — use a pair of parallel sides. In a parallelogram AD ∥ BC, and since E lies on AD produced, the whole line AE is parallel to BC. Taking BE as a transversal cutting these two parallel lines, ∠AEB and ∠CBF are alternate interior angles:

Call this equation (2).

Step 3 — apply the AA criterion. Equations (1) and (2) give two pairs of equal corresponding angles, so:

Hence shown. Check the correspondence: A pairs with C from equation (1), E pairs with B from equation (2), which forces B to pair with F — matching the order △ABE ~ △CFB exactly as asked.

(i) Prove △ABC ~ △AMP

Step 1 — the right angles. We are told the triangles are right angled at B and at M respectively:

Step 2 — the shared angle. Both triangles have a vertex at A, and it is the same angle in each:

Step 3 — apply AA. Two pairs of equal corresponding angles:

(ii) Prove \(\dfrac{CA}{PA}=\dfrac{BC}{MP}\)

Step 4. Corresponding sides of similar triangles are in the same ratio. Under the correspondence A↔A, B↔M, C↔P found in part (i):

Step 5. Taking just the last two of those three equal ratios:

Hence proved.

What we start with. From △ABC ~ △FEG, reading the correspondence A↔F, B↔E, C↔G:

Call these equations (1). Also, since CD bisects ∠ACB and GH bisects ∠EGF, each cuts its angle into two equal halves:

Since the two whole angles are equal by (1), their halves are equal too:

Call these equations (2).

(iii) Prove △DCA ~ △HGF — taken first, because parts (i) and (ii) both lean on it.

In △DCA and △HGF:

from equations (1), and

from equations (2). Two pairs of equal angles, so by AA:

(i) Prove \(\dfrac{CD}{GH}=\dfrac{AC}{FG}\)

From the similarity just established, corresponding sides are proportional. Matching D↔H, C↔G, A↔F:

which is exactly

(ii) Prove △DCB ~ △HGE

In △DCB and △HGE:

from equations (1), and

from equations (2). By AA:

All three shown. The one idea doing the work throughout: equal angles have equal halves, so a pair of corresponding angle bisectors in similar triangles behaves exactly like a pair of corresponding sides.

Step 1 — use the isosceles property. In △ABC we are given \(AB=AC\). Angles opposite equal sides are equal, so:

Call this equation (1).

Step 2 — transfer that angle to point E. This is the step the figure is needed for. E lies on CB produced, so E, B and C are collinear with B between E and C. That means the angle ∠ECF at C, measured between CE and CA, is the same angle as ∠ACB. So from equation (1):

(∠ABD is ∠ABC, since D lies on BC.) Call this equation (2).

Step 3 — use the two perpendiculars. We are given AD ⊥ BC and EF ⊥ AC, which gives a right angle in each triangle:

so

Call this equation (3).

Step 4 — apply the AA criterion. Equations (2) and (3) give two pairs of equal corresponding angles:

Hence proved.

What we are given.

Call this equation (1). The difficulty is that AD and PM are medians, not sides, so equation (1) is not yet in a form any similarity criterion accepts. The fix is to find a smaller pair of triangles in which AD and PM are ordinary sides.

Step 1 — use the medians to halve two sides. Since AD is a median of △ABC, D is the mid-point of BC; since PM is a median of △PQR, M is the mid-point of QR. So:

Step 2 — substitute into equation (1). Replacing BC and QR:

The 2s cancel:

Step 3 — recognise a pair of similar triangles. Those three ratios are precisely the three sides of △ABD against the three sides of △PQM. By the SSS similarity criterion:

Step 4 — extract the angle we actually wanted. Corresponding angles of similar triangles are equal, and B corresponds to Q:

Since D lies on BC and M lies on QR, these are the same as the full angles of the big triangles:

Call this equation (2).

Step 5 — finish on the big triangles with SAS. In △ABC and △PQR we now have two pairs of proportional sides from equation (1):

and their included angle equal from equation (2) — ∠B sits between AB and BC, and ∠Q between PQ and QR. So by the SAS similarity criterion:

Hence shown.

The target \(CA^{2}=CB\cdot CD\) has CA appearing twice, which is the signature of a similarity in which CA is a corresponding side of both triangles. So look for two triangles that both contain CA — here, △ADC and △BAC.

Step 1 — the given pair of angles.

Call this equation (1).

Step 2 — the shared angle. Both triangles have a vertex at C, and since D lies on BC it is the same angle in each:

Call this equation (2).

Step 3 — apply AA. Two pairs of equal corresponding angles, so:

Read the correspondence off equations (1) and (2): D↔A, C↔C, and therefore A↔B.

Step 4 — write the side ratios. Under that correspondence:

Step 5 — cross-multiply.

Hence shown. Notice how the repeated CA in the answer came from CA appearing once in each triangle — whenever a question asks you to prove that some length squared equals a product of two others, that is the structure to hunt for.

This looks like Q12 but is genuinely harder, and the difference is worth noticing before starting. In Q12 the two given sides were AB and BC — and BC is the side the median AD lands on, so halving BC produced a triangle (△ABD) whose three sides were exactly the three given quantities. Here the given sides are AB and AC, the two sides that meet at A, the vertex the median comes from. Halving BC no longer helps directly, so an extra construction is needed.

What we are given.

Call this equation (1).

Step 1 — the construction: double each median. Produce AD to E so that \(AD=DE\), and join CE. Similarly produce PM to N so that \(PM=MN\), and join RN.

Step 2 — show ABEC is a parallelogram. In quadrilateral ABEC the diagonals are BC and AE. D is the mid-point of BC (as AD is a median) and D is also the mid-point of AE (by construction). Since the diagonals bisect each other, ABEC is a parallelogram, and therefore its opposite sides are equal:

By the identical argument, PQNR is a parallelogram and:

Step 3 — rewrite equation (1) using these. Replace AB by CE and PQ by RN in the first ratio:

Step 4 — handle the third ratio. By construction \(AE=2\,AD\) and \(PN=2\,PM\), so:

Combining with equation (1), all three ratios agree:

Step 5 — a first similarity, by SSS. Those are the three sides of △ACE against the three sides of △PRN:

so their corresponding angles are equal, in particular:

Call this equation (2).

Step 6 — repeat on the other half. Running Steps 2–5 with B in place of C gives, in the same way, \(\triangle ABE\sim\triangle PQN\) and hence:

Call this equation (3).

Step 7 — add the two angles. Since E lies on the far side of D, the ray AE is inside ∠BAC and splits it into ∠BAE and ∠CAE. So adding equations (2) and (3):

Call this equation (4).

Step 8 — finish with SAS. In △ABC and △PQR we have from equation (1):

and their included angle equal from equation (4) — ∠A lies between AB and AC, and ∠P between PQ and PR. Therefore by the SAS similarity criterion:

Hence shown.

Step 1 — why the two triangles are similar. Let the tower’s height be \(h\) metres. The pole with its shadow forms one right triangle, and the tower with its shadow forms another. In these two triangles:

because both stand vertically on level ground, and

because the two shadows are cast at the same time, so the sun’s rays strike both at the same angle. (This phrase in the question is not decoration — without it the two triangles need not be similar at all.) By the AA criterion the two triangles are similar.

Step 2 — write the ratio of corresponding sides. Height pairs with height and shadow with shadow:

Step 3 — substitute the numbers.

Step 4 — solve. Simplify the left side first:

Answer: the height of the tower is 42 m.

This is the converse direction of Q12: there we were given the median ratio and had to deduce the similarity; here we are given the similarity and must deduce the median ratio.

What we start with. From △ABC ~ △PQR:

Call this equation (1), and the corresponding angles are equal:

Call this equation (2).

Step 1 — halve the sides the medians land on. Since AD is a median, D is the mid-point of BC; since PM is a median, M is the mid-point of QR:

Step 2 — build a ratio of the halves. Dividing one by the other:

and by equation (1) that equals \(\dfrac{AB}{PQ}\), so:

Call this equation (3). Halving both sides of a ratio leaves the ratio unchanged — that is the only trick being used.

Step 3 — a similarity on the half-triangles, by SAS. In △ABD and △PQM we have two pairs of proportional sides from equation (3) and their included angle equal from equation (2) (∠B lies between AB and BD, and ∠Q between PQ and QM). So:

Step 4 — read off the third ratio. Corresponding sides of these similar triangles are in the same ratio, so including the third pair AD and PM:

Taking the first and last:

Hence proved — corresponding medians of similar triangles are in the same ratio as the corresponding sides, the same fact Example 8(ii) established by a slightly different route.