NCERT Class 10 Mathematics — Triangles, Exercise 6.3. All 16 questions solved.

Exercise 6.3 turns similarity into a working tool. The definition asks for all three pairs of angles to be equal and all three ratios of sides to be equal. For triangles, the criteria of Section 6.4 show that much less is enough:

• AAA, or AA. If two angles of one triangle are equal to two angles of another, the triangles are similar. The third angles then agree automatically, because the angles of each triangle add up to $180^\circ$.
• SSS. If the three sides of one triangle are proportional to the three sides of another, the triangles are similar.
• SAS. If one angle of a triangle equals one angle of another, and the sides including those angles are proportional, the triangles are similar.

The notation carries information of its own. $\triangle ABC \sim \triangle PQR$ means $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$, so $\angle A = \angle P$, $\angle B = \angle Q$, $\angle C = \angle R$ and

$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}.$$

Writing the vertices in a different order makes a different claim, usually a false one.

Key insight. Nearly every proof in this exercise is AA, and the two angles nearly always come from the same short list: a right angle in each triangle, an angle both triangles share, a pair of vertically opposite angles, or alternate angles from a pair of parallel lines. Find two of those, write the vertices in matching order, and the ratio or product of lengths the question wants can be read straight off the similarity.

Question 1

State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Solution. For each pair, use what the figure actually gives. If it gives angles, compare angles. If it gives three sides, compare the three ratios, shortest with shortest and longest with longest. If it gives two sides and an angle, check that the angle is the one between those two sides.

Question 1 (i)

$\triangle ABC$ has $\angle A = 60^\circ$, $\angle B = 80^\circ$, $\angle C = 40^\circ$, and $\triangle PQR$ has $\angle P = 60^\circ$, $\angle Q = 80^\circ$, $\angle R = 40^\circ$. The angles match in the order $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$, so by the AAA criterion $\triangle ABC \sim \triangle PQR$.

Question 1 (ii)

The sides are $AB = 2$, $BC = 2.5$, $CA = 3$ and $PQ = 6$, $QR = 4$, $PR = 5$. Pairing them in order of size,

$$\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}, \qquad \frac{BC}{RP} = \frac{2.5}{5} = \frac{1}{2}, \qquad \frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2}.$$

All three ratios are equal, so the triangles are similar by SSS. The matched sides fix the correspondence. $AB \leftrightarrow QR$ and $BC \leftrightarrow RP$ meet at $B$ and at $R$, so $B \leftrightarrow R$; then $A \leftrightarrow Q$ and $C \leftrightarrow P$. So $\triangle ABC \sim \triangle QRP$, not $\triangle ABC \sim \triangle PQR$.

Question 1 (iii)

The sides are $LM = 2.7$, $MP = 2$, $LP = 3$ and $DE = 4$, $EF = 5$, $DF = 6$. Pairing them in order of size,

$$\frac{MP}{DE} = \frac{2}{4} = 0.5, \qquad \frac{LM}{EF} = \frac{2.7}{5} = 0.54, \qquad \frac{LP}{DF} = \frac{3}{6} = 0.5.$$

The middle ratio is different, so the sides are not proportional. No other matching can do better. If the sides of one triangle are $k$ times those of the other, the shortest side must go with the shortest, and so on. So the triangles are not similar.

Question 1 (iv)

In $\triangle MNL$, $MN = 2.5$, $ML = 5$ and $\angle M = 70^\circ$, the angle between those two sides. In $\triangle PQR$, $QP = 5$, $QR = 10$ and $\angle Q = 70^\circ$, again between the two given sides. So

$$\frac{MN}{QP} = \frac{2.5}{5} = \frac{1}{2} = \frac{5}{10} = \frac{ML}{QR}, \qquad \angle M = \angle Q.$$

Two sides are proportional and the angles they include are equal, so by SAS the triangles are similar. The correspondence is $M \leftrightarrow Q$, $N \leftrightarrow P$ (the other ends of $MN$ and $QP$) and $L \leftrightarrow R$, giving $\triangle MNL \sim \triangle QPR$.

Question 1 (v)

In $\triangle ABC$, $AB = 2.5$, $BC = 3$ and $\angle A = 80^\circ$. In $\triangle DEF$, $DF = 5$, $EF = 6$ and $\angle F = 80^\circ$. The side ratios agree, $\frac{2.5}{5} = \frac{3}{6} = \frac{1}{2}$, but SAS needs the equal angles to be the ones included between the proportional sides. In $\triangle DEF$ they are: $\angle F$ lies between $DF$ and $EF$. In $\triangle ABC$ they are not: the angle between $AB$ and $BC$ is $\angle B$, whereas the $80^\circ$ is at $A$. SAS does not apply, and the pair is not similar.

This is a real “not similar”, not just “cannot tell”. Each triangle is fixed by its data. Solving them with the sine and cosine rules, which come in a later class, gives angles of about $80^\circ$, $44.8^\circ$, $55.2^\circ$ for $\triangle ABC$ and $80^\circ$, $43.8^\circ$, $56.2^\circ$ for $\triangle DEF$. The angles are close, which is why the figure looks convincing, but they are not equal.

Question 1 (vi)

In $\triangle DEF$, $\angle D = 70^\circ$ and $\angle E = 80^\circ$, so $\angle F = 180^\circ – 70^\circ – 80^\circ = 30^\circ$. In $\triangle PQR$, $\angle Q = 80^\circ$ and $\angle R = 30^\circ$, so $\angle P = 180^\circ – 80^\circ – 30^\circ = 70^\circ$. Hence $\angle D = \angle P$, $\angle E = \angle Q$ and $\angle F = \angle R$, and by the AA criterion $\triangle DEF \sim \triangle PQR$.

Question 2

In Fig. 6.35, $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^\circ$ and $\angle CDO = 70^\circ$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.

$D$ and $C$ lie on the upper of two lines drawn across the figure, $A$ and $B$ on the lower one, and the segments $AC$ and $BD$ cross at $O$.

Solution.

$\angle DOC$. $D$, $O$ and $B$ lie on one straight line, so $\angle DOC$ and $\angle BOC$ form a linear pair:

$$\angle DOC = 180^\circ – 125^\circ = 55^\circ.$$

$\angle DCO$. The angles of $\triangle ODC$ add up to $180^\circ$:

$$\angle DCO = 180^\circ – \angle CDO – \angle DOC = 180^\circ – 70^\circ – 55^\circ = 55^\circ.$$

As a check, $\angle BOC$ is an exterior angle of $\triangle ODC$, so it equals the sum of the two interior opposite angles: $70^\circ + 55^\circ = 125^\circ$.

$\angle OAB$. The similarity $\triangle ODC \sim \triangle OBA$ pairs $O \leftrightarrow O$, $D \leftrightarrow B$, $C \leftrightarrow A$. So $\angle OAB$ corresponds to $\angle OCD$:

$$\angle OAB = \angle OCD = 55^\circ.$$

Question 3

Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for two triangles, show that $\dfrac{OA}{OC} = \dfrac{OB}{OD}$.

Solution. The ratio pairs $OA$ with $OC$ and $OB$ with $OD$, so we want a similarity with $A \leftrightarrow C$ and $B \leftrightarrow D$. That means comparing $\triangle OAB$ with $\triangle OCD$.

In $\triangle OAB$ and $\triangle OCD$, $\angle OAB = \angle OCD$, because they are alternate angles ($AB \parallel DC$, with $AC$ as the transversal). Likewise $\angle OBA = \angle ODC$ are alternate angles, with $BD$ as the transversal. (The angles at $O$ are also equal, being vertically opposite, but two pairs are enough.)

So by the AA criterion, $\triangle OAB \sim \triangle OCD$, with $O \leftrightarrow O$, $A \leftrightarrow C$, $B \leftrightarrow D$. Corresponding sides are proportional:

$$\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}.$$

This is Exercise 6.2, question 9 again. There the Basic Proportionality Theorem needed an extra line through $O$. Similarity needs no construction.

Question 4

In Fig. 6.36, $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\triangle PQS \sim \triangle TQR$.

$S$ lies on $QR$, and $T$ lies on $QP$ produced beyond $P$. $PS$, $PR$ and $TR$ are drawn. $\angle 1$ is the angle at $Q$ of $\triangle PQR$, and $\angle 2$ is its angle at $R$, between $RQ$ and $RP$.

Solution. In the target $\triangle PQS \sim \triangle TQR$, $Q$ corresponds to $Q$, and the angle at $Q$ is shared by both triangles. So SAS is the natural route, and for it we need $\dfrac{QP}{QT} = \dfrac{QS}{QR}$. The given ratio involves $PR$ rather than $PQ$, and $\angle 1 = \angle 2$ is what lets us swap one for the other.

In $\triangle PQR$, $\angle PQR = \angle PRQ$ (this is $\angle 1 = \angle 2$), so the sides opposite those angles are equal: $PR = PQ$. Substituting in the given ratio,

$$\frac{QR}{QS} = \frac{QT}{PQ} \quad\Longrightarrow\quad \frac{PQ}{TQ} = \frac{QS}{QR}.$$

Now compare $\triangle PQS$ and $\triangle TQR$. We have $\dfrac{PQ}{TQ} = \dfrac{QS}{QR}$, and the angle included between these sides, $\angle PQS = \angle TQR$, is $\angle 1$ itself, common to both triangles. By SAS, $\triangle PQS \sim \triangle TQR$, with $P \leftrightarrow T$, $Q \leftrightarrow Q$, $S \leftrightarrow R$.

Question 5

$S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle RTS$. Show that $\triangle RPQ \sim \triangle RTS$.

Solution. Compare $\triangle RPQ$ and $\triangle RTS$. $\angle RPQ = \angle RTS$ is given. $\angle PRQ = \angle TRS$, because both are the angle at $R$: $S$ lies on $RP$ and $T$ lies on $RQ$. By AA, $\triangle RPQ \sim \triangle RTS$, with $R \leftrightarrow R$, $P \leftrightarrow T$, $Q \leftrightarrow S$.

Notice that $P$ corresponds to $T$ even though $T$ lies on $QR$. The segment $ST$ is not parallel to $PQ$: it cuts the two sides “the other way round”, so do not reach for the Basic Proportionality Theorem here.

Question 6

In Fig. 6.37, if $\triangle ABE \cong \triangle ACD$, show that $\triangle ADE \sim \triangle ABC$.

$D$ lies on $AB$ and $E$ on $AC$; $DE$, $BE$ and $CD$ are drawn.

Solution. Congruence hands us equal sides, and the angle at $A$ is shared, so SAS is the route.

$\triangle ABE \cong \triangle ACD$ pairs $A \leftrightarrow A$, $B \leftrightarrow C$, $E \leftrightarrow D$, so corresponding parts are equal (CPCT):

$$AB = AC, \qquad AE = AD.$$

Hence

$$\frac{AD}{AB} = \frac{AE}{AC},$$

since the two numerators are equal and so are the two denominators. In $\triangle ADE$ and $\triangle ABC$, the included angle $\angle DAE = \angle BAC$ is common. By SAS, $\triangle ADE \sim \triangle ABC$, with $A \leftrightarrow A$, $D \leftrightarrow B$, $E \leftrightarrow C$.

Question 7

In Fig. 6.38, altitudes $AD$ and $CE$ of $\triangle ABC$ intersect each other at the point $P$. Show that:

(i) $\triangle AEP \sim \triangle CDP$

(ii) $\triangle ABD \sim \triangle CBE$

(iii) $\triangle AEP \sim \triangle ADB$

(iv) $\triangle PDC \sim \triangle BEC$

$D$ is the foot of the altitude from $A$ on $BC$, $E$ is the foot of the altitude from $C$ on $AB$, and $P$ is where the two altitudes cross.

Solution. All four parts are AA. In each one, the first pair of angles is a pair of right angles at the feet of the altitudes, since $\angle ADB = \angle ADC = 90^\circ$ and $\angle AEC = \angle BEC = 90^\circ$. Only the second pair changes.

(i) In $\triangle AEP$ and $\triangle CDP$, $\angle AEP = \angle CDP = 90^\circ$, and $\angle APE = \angle CPD$ because they are vertically opposite angles. By AA, $\triangle AEP \sim \triangle CDP$.

(ii) In $\triangle ABD$ and $\triangle CBE$, $\angle ADB = \angle CEB = 90^\circ$, and $\angle ABD = \angle CBE$, both being $\angle B$ of $\triangle ABC$. By AA, $\triangle ABD \sim \triangle CBE$.

(iii) In $\triangle AEP$ and $\triangle ADB$, $\angle AEP = \angle ADB = 90^\circ$, and $\angle PAE = \angle BAD$, the same angle at $A$, because $P$ lies on $AD$ and $E$ lies on $AB$. By AA, $\triangle AEP \sim \triangle ADB$.

(iv) In $\triangle PDC$ and $\triangle BEC$, $\angle PDC = \angle BEC = 90^\circ$, and $\angle PCD = \angle BCE$, the same angle at $C$, because $P$ lies on $CE$ and $D$ lies on $CB$. By AA, $\triangle PDC \sim \triangle BEC$.

Question 8

$E$ is a point on the side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Show that $\triangle ABE \sim \triangle CFB$.

Solution. Compare $\triangle ABE$ and $\triangle CFB$.

$\angle BAE = \angle FCB$. $\angle BAE$ is $\angle A$ of the parallelogram, since $E$ lies on $AD$ produced, and $\angle FCB$ is $\angle C$, since $F$ lies on $CD$. Opposite angles of a parallelogram are equal.

$\angle AEB = \angle CBF$. $AE$ lies along $AD$, which is parallel to $BC$, and $BE$ is a transversal, so these are alternate angles.

By AA, $\triangle ABE \sim \triangle CFB$, with $A \leftrightarrow C$, $B \leftrightarrow F$, $E \leftrightarrow B$.

Question 9

In Fig. 6.39, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that:

(i) $\triangle ABC \sim \triangle AMP$

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

$A$, $B$ and $P$ lie on one straight line, with $C$ above $B$. $M$ lies on $AC$, and $PM \perp AC$.

Solution.

(i) In $\triangle ABC$ and $\triangle AMP$, $\angle ABC = \angle AMP = 90^\circ$, as given. $\angle BAC = \angle MAP$, because both are the angle at $A$: $B$ lies on $AP$ and $M$ lies on $AC$. By AA, $\triangle ABC \sim \triangle AMP$, with $A \leftrightarrow A$, $B \leftrightarrow M$, $C \leftrightarrow P$.

(ii) Corresponding sides of similar triangles are proportional, so

$$\frac{AB}{AM} = \frac{BC}{MP} = \frac{CA}{PA}.$$

The last two ratios give $\dfrac{CA}{PA} = \dfrac{BC}{MP}$.

Question 10

$CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that:

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

(ii) $\triangle DCB \sim \triangle HGE$

(iii) $\triangle DCA \sim \triangle HGF$

Solution. Read the correspondence carefully first. $\triangle ABC \sim \triangle FEG$ means $A \leftrightarrow F$, $B \leftrightarrow E$, $C \leftrightarrow G$, so

$$\angle A = \angle F, \qquad \angle B = \angle E, \qquad \angle ACB = \angle FGE.$$

Each bisector cuts its angle into two equal halves, and the whole angles are equal, so all four half-angles are equal:

$$\angle ACD = \angle DCB = \tfrac{1}{2}\angle ACB = \tfrac{1}{2}\angle FGE = \angle FGH = \angle HGE.$$

(i) Compare $\triangle DCA$ and $\triangle HGF$. $\angle DAC = \angle HFG$, since these are $\angle A$ and $\angle F$. $\angle DCA = \angle HGF$, since these are half-angles. By AA, $\triangle DCA \sim \triangle HGF$, with $D \leftrightarrow H$, $C \leftrightarrow G$, $A \leftrightarrow F$. Corresponding sides are proportional, so

$$\frac{DC}{HG} = \frac{CA}{GF}, \quad\text{that is,}\quad \frac{CD}{GH} = \frac{AC}{FG}.$$

(ii) Compare $\triangle DCB$ and $\triangle HGE$. $\angle DBC = \angle HEG$, since these are $\angle B$ and $\angle E$. $\angle DCB = \angle HGE$, since these are half-angles. By AA, $\triangle DCB \sim \triangle HGE$.

(iii) This is the similarity $\triangle DCA \sim \triangle HGF$ already proved by AA in part (i).

Question 11

In Fig. 6.40, $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, prove that $\triangle ABD \sim \triangle ECF$.

$D$ is the foot of the perpendicular from $A$ to $BC$. $E$ lies on $CB$ extended beyond $B$ (dotted in the figure), and $F$ is the foot of the perpendicular from $E$ to $AC$.

Solution. Compare $\triangle ABD$ and $\triangle ECF$.

$\angle ADB = \angle EFC = 90^\circ$, since $AD \perp BC$ and $EF \perp AC$.

$\angle ABD = \angle ECF$. Because $AB = AC$, the angles opposite these sides are equal, so $\angle ABC = \angle ACB$. Now $\angle ABD$ is $\angle ABC$, as $D$ lies on $BC$. And $\angle ECF$ is $\angle ACB$, as $E$ lies on line $CB$ on the same side of $C$ as $B$, and $F$ lies on $CA$.

By AA, $\triangle ABD \sim \triangle ECF$, with $A \leftrightarrow E$, $B \leftrightarrow C$, $D \leftrightarrow F$.

Question 12

Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \triangle PQR$.

$D$ is the mid-point of $BC$ and $M$ is the mid-point of $QR$.

Solution. We are given

$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}.$$

Nothing yet tells us about the angles of $\triangle ABC$ and $\triangle PQR$. So first find a similar pair among the smaller triangles $\triangle ABD$ and $\triangle PQM$, where all three sides are available.

Since $D$ and $M$ are mid-points, $BD = \frac{1}{2}BC$ and $QM = \frac{1}{2}QR$, so

$$\frac{BD}{QM} = \frac{\frac{1}{2}BC}{\frac{1}{2}QR} = \frac{BC}{QR}.$$

Therefore

$$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM},$$

and $\triangle ABD \sim \triangle PQM$ by SSS, with $A \leftrightarrow P$, $B \leftrightarrow Q$, $D \leftrightarrow M$. In particular $\angle ABD = \angle PQM$, that is, $\angle B = \angle Q$.

Now compare $\triangle ABC$ and $\triangle PQR$. $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ is given, and the angles included between those sides are $\angle B = \angle Q$. By SAS, $\triangle ABC \sim \triangle PQR$.

Question 13

$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.

Solution. Write the target as a proportion: $CA^2 = CB \cdot CD$ is the same as $\dfrac{CD}{CA} = \dfrac{CA}{CB}$. So we want one triangle containing $CD$ and $CA$, and another containing $CA$ and $CB$: $\triangle ADC$ and $\triangle BAC$.

In $\triangle ADC$ and $\triangle BAC$, $\angle ADC = \angle BAC$ is given. $\angle ACD = \angle BCA$, because both are the angle at $C$ ($D$ lies on $CB$). By AA, $\triangle ADC \sim \triangle BAC$, with $A \leftrightarrow B$, $D \leftrightarrow A$, $C \leftrightarrow C$.

Corresponding sides are proportional:

$$\frac{AD}{BA} = \frac{DC}{AC} = \frac{CA}{CB}.$$

The last two give $\dfrac{CD}{CA} = \dfrac{CA}{CB}$, and cross-multiplying gives $CA^2 = CB \cdot CD$.

Question 14

Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC \sim \triangle PQR$.

Solution. This looks like question 12, but now the two known sides meet at the vertex the median starts from. So none of the smaller triangles has three known sides. The hint printed with the textbook’s answers builds one: produce $AD$ to $E$ so that $DE = AD$, produce $PM$ to $N$ so that $MN = PM$, and join $EC$ and $NR$.

Step 1: $EC = AB$ and $NR = PQ$. In $\triangle ABD$ and $\triangle ECD$, $BD = DC$ because $D$ is the mid-point, $AD = ED$ by construction, and $\angle ADB = \angle EDC$ because they are vertically opposite. By SAS congruence, $\triangle ABD \cong \triangle ECD$, so $AB = EC$ and $\angle BAD = \angle CED$. In the same way $\triangle PQM \cong \triangle NRM$, so $PQ = NR$ and $\angle QPM = \angle RNM$.

Step 2: $\triangle AEC \sim \triangle PNR$. We are given $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$. Replace $AB$ by $EC$ and $PQ$ by $NR$, and use $AE = 2AD$ and $PN = 2PM$:

$$\frac{EC}{NR} = \frac{AC}{PR} = \frac{2AD}{2PM} = \frac{AE}{PN}.$$

By SSS, $\triangle AEC \sim \triangle PNR$, with $A \leftrightarrow P$, $E \leftrightarrow N$, $C \leftrightarrow R$. Hence $\angle CAE = \angle RPN$ and $\angle AEC = \angle PNR$.

Step 3: $\angle BAC = \angle QPR$. From step 1, $\angle BAD = \angle CED = \angle AEC$ and $\angle QPM = \angle RNM = \angle PNR$. Step 2 says these are equal, so $\angle BAD = \angle QPM$. Step 2 also gives $\angle CAD = \angle CAE = \angle RPN = \angle RPM$. Adding,

$$\angle BAC = \angle BAD + \angle DAC = \angle QPM + \angle MPR = \angle QPR.$$

Step 4. In $\triangle ABC$ and $\triangle PQR$, $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$ is given, and the angles included between those sides are equal, $\angle BAC = \angle QPR$. By SAS, $\triangle ABC \sim \triangle PQR$.

Question 15

A vertical pole of length $6$ m casts a shadow $4$ m long on the ground and at the same time a tower casts a shadow $28$ m long. Find the height of the tower.

Solution. Let the pole be $AB$, with shadow $BC$, and let the tower be $PQ$, of height $h$ m, with shadow $QR$. Both stand vertically, so $\angle ABC = \angle PQR = 90^\circ$. The shadows are cast at the same time, so the sun’s rays make the same angle with the ground in both cases: $\angle ACB = \angle PRQ$. By AA, $\triangle ABC \sim \triangle PQR$, with $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$. Hence

$$\frac{AB}{PQ} = \frac{BC}{QR} \quad\Longrightarrow\quad \frac{6}{h} = \frac{4}{28} \quad\Longrightarrow\quad h = \frac{6 \times 28}{4} = 42.$$

As a check, the tower’s shadow is $7$ times the pole’s, so the tower must be $7$ times as tall: $7 \times 6 = 42$ m.

Question 16

If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectively where $\triangle ABC \sim \triangle PQR$, prove that $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$.

Solution. This is question 12 run the other way. Now the similarity is given, and we must show that the medians are in the same ratio as the sides.

$\triangle ABC \sim \triangle PQR$ gives $\angle B = \angle Q$ and

$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM},$$

using $BC = 2BD$ and $QR = 2QM$, since a median ends at the mid-point of the opposite side. So in $\triangle ABD$ and $\triangle PQM$, $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$, and the included angles are equal, $\angle ABD = \angle PQM$. By SAS, $\triangle ABD \sim \triangle PQM$, and therefore

$$\frac{AB}{PQ} = \frac{AD}{PM}.$$

Common mistakes

• Question 1(ii), writing $\triangle ABC \sim \triangle PQR$. The triangles are similar, but that statement pairs $AB = 2$ with $PQ = 6$ and $BC = 2.5$ with $QR = 4$, ratios of $\frac{1}{3}$ and $\frac{5}{8}$. Matching the sides by size gives $A \leftrightarrow Q$, $B \leftrightarrow R$, $C \leftrightarrow P$, so the correct statement is $\triangle ABC \sim \triangle QRP$.
• Question 1(v), calling the pair similar by SAS. Two proportional sides and an equal angle are not enough unless the angle lies between those sides. In $\triangle ABC$ the $80^\circ$ is at $A$, not between $AB$ and $BC$.
• Question 1(iii), pairing sides in the order they happen to be labelled. Sort both triangles’ sides and pair shortest with shortest. The ratios then come out as $0.5$, $0.54$, $0.5$. They are close, but one mismatch is enough to rule similarity out.
• Question 2, taking $\angle OAB$ as equal to $\angle ODC$. The similarity $\triangle ODC \sim \triangle OBA$ pairs $D$ with $B$ and $C$ with $A$, so $\angle OAB$ matches $\angle OCD = 55^\circ$. Reading the letters in the wrong order gives $70^\circ$.
• Question 4, trying to use $\frac{QR}{QS} = \frac{QT}{PR}$ as it stands. $PR$ is not a side of either triangle being compared. The condition $\angle 1 = \angle 2$ is there precisely to replace $PR$ by $PQ$.
• Question 5, writing $\triangle RPQ \sim \triangle RST$. The equal angles are $\angle P$ and $\angle RTS$, so $P$ goes with $T$, not with $S$. $ST$ is not parallel to $PQ$.
• Questions 12 and 14, applying SAS to the large triangles straight away. In question 12 the proportional sides $AB$ and $BC$ include $\angle B$, but nothing says $\angle B = \angle Q$ until $\triangle ABD \sim \triangle PQM$ proves it. In question 14 no small triangle has three known sides until $E$ and $N$ are added.
• Question 15, skipping the reason for similarity. “At the same time” is what makes the sun’s rays meet the ground at the same angle for both. That equal angle, together with the right angles, is what gives AA.

Practise next

• Exercise 6.2 — the Basic Proportionality Theorem; its question 9 is the trapezium result of question 3 proved without similar triangles.
• Chapter 9, Exercise 9.1 — heights and distances, where shadow problems like question 15 return and are solved with trigonometric ratios.