NCERT Class 10 Exercise 6.3 Solutions: Similar Triangles

Exer­cise 6.3 of the NCERT Class 10 chap­ter Tri­an­gles is where the three sim­i­lar­ity cri­te­ria get used on real fig­ures. It has six­teen ques­tions: read­ing sim­i­lar pairs off a fig­ure, find­ing angles, and prov­ing that two tri­an­gles are sim­i­lar when medi­ans, alti­tudes, angle bisec­tors, a par­al­lel­o­gram or a trapez­ium are involved. It ends with a shadow prob­lem. Every solu­tion below is writ­ten out in full, the way you would set it out in an exam, and every numer­i­cal check has been worked through again.

The meth­ods you need#

The three sim­i­lar­ity cri­te­ria#

• AA (angle–angle): if two angles of one tri­an­gle are equal to two angles of another, the tri­an­gles are sim­i­lar. The third angles are then equal too, because every tri­an­gle's angles add up to $180^\circ$.
• SSS (side–side–side): if all three sides of one tri­an­gle are pro­por­tional to the three sides of another, the tri­an­gles are sim­i­lar.
• SAS (side–angle–side): if one angle of a tri­an­gle equals one angle of another, and the sides includ­ing those angles are pro­por­tional, the tri­an­gles are sim­i­lar.

How to write a sim­i­lar­ity proof#

1. Name the two tri­an­gles you will com­pare, and mark what you know about each on the fig­ure.
2. Look for free facts: a com­mon angle, ver­ti­cally oppo­site angles, alter­nate angles between par­al­lel lines, right angles from per­pen­dic­u­lars, and equal base angles of an isosce­les tri­an­gle.
3. State the cri­te­rion (AA, SSS or SAS) and write the sim­i­lar­ity with match­ing ver­tices in the same order, for exam­ple $\triangle ABC \sim \triangle PQR$ when $\angle A = \angle P$, $\angle B = \angle Q$ and $\angle C = \angle R$.
4. Read off the equal ratios of cor­re­spond­ing sides, and fin­ish with the result you were asked to prove.

The order of the let­ters mat­ters. In $\triangle ABC \sim \triangle PQR$ the side $AB$ cor­re­sponds to $PQ$, $BC$ to $QR$ and $CA$ to $RP$. If you write the let­ters in the wrong order, you will read off the wrong ratios.

Solu­tions to Exer­cise 6.3#

Ques­tion 1#

State which pairs of tri­an­gles in Fig. 6.34 are sim­i­lar. Write the sim­i­lar­ity cri­te­rion you used and write the sim­i­lar pairs in sym­bolic form.

The six pairs of tri­an­gles for this ques­tion are drawn in Fig. 6.34 of the NCERT text­book, and their angles and sides are marked only on that fig­ure. Keep the text­book open and read the val­ues for each pair from it. Then decide each pair as fol­lows.

Step 1. Match the marked data to a test.

• AA: if two angles are marked in each tri­an­gle, com­pare them. If two pairs are equal, the tri­an­gles are sim­i­lar. If only two angles are marked, find the third angle using the angle sum of $180^\circ$ before you com­pare.
• SSS: if three sides are marked in each tri­an­gle, pair the short­est with the short­est, the mid­dle with the mid­dle and the longest with the longest, and divide. If all three ratios are equal, the tri­an­gles are sim­i­lar. If even one ratio is dif­fer­ent, they are not.
• SAS: if two sides and one angle are marked in each tri­an­gle, first check that the angle lies between the two sides. Then check that the angles are equal and that the ratios of the two pairs of sides are equal. If the angle is not the included angle, SAS can­not be used.

Step 2. Write the cor­re­spon­dence. List the ver­tices in match­ing order, so that equal angles sit in the same posi­tion on both sides of $\sim$. If $\angle A = \angle Q$, $\angle B = \angle R$ and $\angle C = \angle P$, write $\triangle ABC \sim \triangle QRP$, not $\triangle ABC \sim \triangle PQR$. For SSS, the ver­tex oppo­site the short­est side matches the ver­tex oppo­site the short­est side, and so on.

Illus­tra­tion (hypo­thet­i­cal val­ues, not the text­book's). Sup­pose $\triangle XYZ$ has $XY = 3$, $YZ = 4$, $ZX = 5$, and $\triangle KLM$ has $KL = 10$, $LM = 6$, $MK = 8$. Sort each set of sides and divide:

$$\displaystyle \frac{XY}{LM} = \frac{3}{6} = \frac12, \qquad \frac{YZ}{MK} = \frac{4}{8} = \frac12, \qquad \frac{ZX}{KL} = \frac{5}{10} = \frac12.$$

All three ratios are equal, so the tri­an­gles are sim­i­lar by SSS. Side $XY$ matches $LM$ and side $YZ$ matches $MK$, so ver­tex $Y$ (com­mon to $XY$ and $YZ$) matches ver­tex $M$ (com­mon to $LM$ and $MK$). Con­tin­u­ing the same way, $X \leftrightarrow L$ and $Z \leftrightarrow K$, so we write $\triangle XYZ \sim \triangle LMK$.

A sec­ond illus­tra­tion, also hypo­thet­i­cal: if one tri­an­gle has $\angle U = 50^\circ$ with sides $UV = 2$ and $VW = 3$, the $50^\circ$ angle is not between those two sides. No mat­ter what the other tri­an­gle shows, SAS can­not be applied, so you can­not con­clude that the tri­an­gles are sim­i­lar.

Answer: for each pair in Fig. 6.34, name the cri­te­rion whose con­di­tions the marked data fully sat­is­fies (AA, SSS or SAS) and write the pair with the ver­tices in cor­re­spond­ing order. If no cri­te­rion is fully sat­is­fied, for exam­ple because one side ratio dif­fers or the marked angle is not the included angle, write that the pair is not sim­i­lar.

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 $\angle OAB$.

In the fig­ure, the seg­ments $AC$ and $BD$ cross at $O$.

Step 1. $A$, $O$, $C$ lie on a straight line, so $\angle BOC$ and $\angle AOB$ form a lin­ear pair:

$$\angle AOB = 180^\circ - \angle BOC = 180^\circ - 125^\circ = 55^\circ.$$

Step 2. $\angle DOC$ and $\angle AOB$ are ver­ti­cally oppo­site angles, so

$$\angle DOC = \angle AOB = 55^\circ.$$

Step 3. Use the angle sum prop­erty in $\triangle ODC$:

$$\begin{aligned} \angle DCO &= 180^\circ - \angle CDO - \angle DOC \\ &= 180^\circ - 70^\circ - 55^\circ = 55^\circ. \end{aligned}$$

Step 4. In $\triangle ODC \sim \triangle OBA$ the ver­tices cor­re­spond as $O \leftrightarrow O$, $D \leftrightarrow B$, $C \leftrightarrow A$. So $\angle OAB$ cor­re­sponds to $\angle OCD$:

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

Check: the sim­i­lar­ity also says $\angle DOC = \angle BOA$, and both are $55^\circ$, so the work­ing is con­sis­tent.

Answer: $\angle DOC = 55^\circ$, $\angle DCO = 55^\circ$, $\angle OAB = 55^\circ$.

Ques­tion 3#

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

Proof. Com­pare $\triangle OAB$ and $\triangle OCD$.

• $\angle AOB = \angle COD$ (ver­ti­cally oppo­site angles).
• $\angle OAB = \angle OCD$ (alter­nate angles, since $AB \parallel DC$ and $AC$ is a trans­ver­sal).

So by the AA cri­te­rion,

$$\triangle OAB \sim \triangle OCD.$$

With the cor­re­spon­dence $O \leftrightarrow O$, $A \leftrightarrow C$, $B \leftrightarrow D$, the cor­re­spond­ing sides are pro­por­tional:

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

Answer: $\displaystyle \frac{OA}{OC} = \frac{OB}{OD}$, as required.

Check with coor­di­nates: take $A(0, 0)$, $B(6, 0)$, $C(4, 3)$, $D(1, 3)$, so $AB = 6$ and $DC = 3$. The diag­o­nals meet at $\displaystyle O(\tfrac{8}{3}, 2)$, and both $\displaystyle \frac{OA}{OC}$ and $\displaystyle \frac{OB}{OD}$ come out as $\displaystyle 2 = \frac{AB}{DC}$.

Why this works: the diag­o­nals of a trapez­ium divide each other in the ratio of the par­al­lel sides.

Ques­tion 4#

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

In the fig­ure, $S$ lies on $QR$, $T$ lies on $QP$ pro­duced beyond $P$, and $\angle 1$ and $\angle 2$ are the angles of $\triangle PQR$ at $Q$ and at $R$.

Step 1. In $\triangle PQR$, $\angle PQR = \angle 1$ and $\angle PRQ = \angle 2$, and $\angle 1 = \angle 2$. Sides oppo­site equal angles are equal, so

$$PQ = PR.$$

Step 2. Replace $PR$ by $PQ$ in the given ratio, then take rec­i­p­ro­cals:

$$\displaystyle \frac{QR}{QS} = \frac{QT}{PQ} \implies \frac{QS}{QR} = \frac{PQ}{QT}, \quad \text{that is,} \quad \frac{PQ}{TQ} = \frac{QS}{QR}.$$

Step 3. $P$ lies on $QT$ and $S$ lies on $QR$, so $\angle PQS$ and $\angle TQR$ are the same angle:

$$\angle PQS = \angle TQR = \angle Q \quad (\text{common angle}).$$

Step 4. In $\triangle PQS$ and $\triangle TQR$, the sides includ­ing $\angle Q$ are pro­por­tional and $\angle Q$ is com­mon. So by the SAS cri­te­rion,

$$\triangle PQS \sim \triangle TQR.$$

Answer: $\triangle PQS \sim \triangle TQR$ (SAS), with $P \leftrightarrow T$, $Q \leftrightarrow Q$, $S \leftrightarrow R$.

Check with the fig­ure's num­bers: with $PQ = PR = 5$, $QR = 6$, $QS = 4$ and $QT = 7.5$, we get $\displaystyle \frac{QR}{QS} = \frac{6}{4} = 1.5$ and $\displaystyle \frac{QT}{PR} = \frac{7.5}{5} = 1.5$. Also $\displaystyle \frac{PQ}{TQ} = \frac{5}{7.5} = \frac23$ and $\displaystyle \frac{QS}{QR} = \frac46 = \frac23$, and the third sides $PS$ and $TR$ are in the same ratio $\displaystyle \frac23$.

Why this works: the given ratio men­tions $PR$, which is not a side of $\triangle PQS$. The equal angles $\angle 1 = \angle 2$ let you replace $PR$ with $PQ$, which is a side of that tri­an­gle.

Ques­tion 5#

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

Proof. Com­pare $\triangle RPQ$ and $\triangle RTS$.

• $\angle PRQ = \angle TRS$, because $S$ lies on $RP$ and $T$ lies on $RQ$, so both are the angle at $R$ (com­mon angle).
• $\angle RPQ = \angle RTS$ (given).

So by the AA cri­te­rion, $\triangle RPQ \sim \triangle RTS$.

Answer: $\triangle RPQ \sim \triangle RTS$ (AA), with $R \leftrightarrow R$, $P \leftrightarrow T$, $Q \leftrightarrow S$.

Check: if $\angle P = 50^\circ$ and $\angle R = 70^\circ$, then $\angle Q = 60^\circ$. In $\triangle RTS$, $\angle RTS = 50^\circ$ and $\angle R = 70^\circ$, so $\angle RST = 60^\circ = \angle Q$. All three angles match.

Ques­tion 6#

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

In the fig­ure, $D$ lies on $AB$ and $E$ lies on $AC$.

Step 1. In $\triangle ABE \cong \triangle ACD$ the ver­tices cor­re­spond as $A \leftrightarrow A$, $B \leftrightarrow C$, $E \leftrightarrow D$. Cor­re­spond­ing parts of con­gru­ent tri­an­gles are equal (CPCT), so

$$AB = AC \quad \text{and} \quad AE = AD.$$

Step 2. Divide the sec­ond equal­ity by the first:

$$\displaystyle \frac{AD}{AB} = \frac{AE}{AC}.$$

Step 3. $\angle DAE = \angle BAC$, because both are the angle at $A$ (com­mon angle).

Step 4. In $\triangle ADE$ and $\triangle ABC$, the sides includ­ing $\angle A$ are pro­por­tional and $\angle A$ is com­mon. So by SAS, $\triangle ADE \sim \triangle ABC$.

Answer: $\triangle ADE \sim \triangle ABC$ (SAS).

Check with num­bers: take $AB = AC = 5$ and $AD = AE = 2$, with $A(0, 0)$, $B(4.330, -2.5)$, $C(4.330, 2.5)$, $D(1.732, -1)$ and $E(1.732, 1)$. Then $\displaystyle \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} = \frac{2}{5} = 0.4$.

Ques­tion 7#

In Fig. 6.38, alti­tudes $AD$ and $CE$ of $\triangle ABC$ inter­sect each other at $P$. Show that: (i) $\triangle AEP \sim \triangle CDP$ (ii) $\triangle ABD \sim \triangle CBE$ (iii) $\triangle AEP \sim \triangle ADB$ (iv) $\triangle PDC \sim \triangle BEC$.

$AD \perp BC$ with $D$ on $BC$, and $CE \perp AB$ with $E$ on $AB$. So $\angle ADB = \angle ADC = 90^\circ$ and $\angle AEC = \angle BEC = 90^\circ$.

(i) In $\triangle AEP$ and $\triangle CDP$:

• $\angle AEP = \angle CDP = 90^\circ$;
• $\angle APE = \angle CPD$ (ver­ti­cally oppo­site angles).

By AA, $\triangle AEP \sim \triangle CDP$.

(ii) In $\triangle ABD$ and $\triangle CBE$:

• $\angle ADB = \angle CEB = 90^\circ$;
• $\angle ABD = \angle CBE$ (both are $\angle B$, a com­mon angle).

By AA, $\triangle ABD \sim \triangle CBE$.

(iii) In $\triangle AEP$ and $\triangle ADB$:

• $\angle AEP = \angle ADB = 90^\circ$;
• $\angle PAE = \angle BAD$ ($P$ lies on $AD$ and $E$ lies on $AB$, so this is a com­mon angle).

By AA, $\triangle AEP \sim \triangle ADB$.

(iv) In $\triangle PDC$ and $\triangle BEC$:

• $\angle PDC = \angle BEC = 90^\circ$;
• $\angle PCD = \angle BCE$ ($P$ lies on $CE$ and $D$ lies on $CB$, so this is a com­mon angle).

By AA, $\triangle PDC \sim \triangle BEC$.

Answer: all four sim­i­lar­i­ties hold by the AA cri­te­rion, as shown above.

Check with coor­di­nates: for $A(0, 4)$, $B(-2, 0)$, $C(3, 0)$, the feet of the alti­tudes are $D(0, 0)$ and $E(-1, 2)$, and the alti­tudes meet at $P(0, 1.5)$. This point also lies on the alti­tude from $B$, so it is the ortho­cen­tre. In every one of the four pairs, the three side ratios come out equal. In (ii), for exam­ple, $\displaystyle \frac{AB}{CB} = \frac{BD}{BE} = \frac{AD}{CE} = \frac{2}{\sqrt5} \approx 0.894$.

Ques­tion 8#

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

Step 1. $E$ lies on $AD$ pro­duced, so $\angle BAE = \angle BAD$. $F$ lies on $CD$, so $\angle BCF = \angle BCD$. Oppo­site angles of a par­al­lel­o­gram are equal, so

$$\angle BAE = \angle BAD = \angle BCD = \angle BCF.$$

Step 2. $AE \parallel BC$ (the line $AD$ is par­al­lel to $BC$), and $EB$ is a trans­ver­sal. So the alter­nate angles are equal:

$$\angle AEB = \angle EBC = \angle FBC.$$

Step 3. In $\triangle ABE$ and $\triangle CFB$, $\angle A = \angle C$ and $\angle E = \angle B$. By AA,

$$\triangle ABE \sim \triangle CFB.$$

Answer: $\triangle ABE \sim \triangle CFB$ (AA), with $A \leftrightarrow C$, $B \leftrightarrow F$, $E \leftrightarrow B$.

Check with coor­di­nates: take $A(0, 0)$, $B(4, 0)$, $C(6, 3)$, $D(2, 3)$. Pro­duce $AD$ to $E(4, 6)$. The line $BE$ is $x = 4$, and it meets $DC$ at $F(4, 3)$. Then $\displaystyle \frac{AB}{CF} = \frac{4}{2}$, $\displaystyle \frac{BE}{FB} = \frac{6}{3}$ and $\displaystyle \frac{EA}{BC} = \frac{\sqrt{52}}{\sqrt{13}}$, and all three ratios equal $2$.

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) $\displaystyle \frac{CA}{PA} = \frac{BC}{MP}$.

(i) In $\triangle ABC$ and $\triangle AMP$:

• $\angle ABC = \angle AMP = 90^\circ$ (given);
• $\angle BAC = \angle MAP$ (com­mon angle at $A$).

By AA, $\triangle ABC \sim \triangle AMP$.

(ii) From (i), with $A \leftrightarrow A$, $B \leftrightarrow M$, $C \leftrightarrow P$, the cor­re­spond­ing sides are pro­por­tional:

$$\displaystyle \frac{AB}{AM} = \frac{BC}{MP} = \frac{CA}{PA}, \quad \text{so} \quad \frac{CA}{PA} = \frac{BC}{MP}.$$

Answer: (i) $\triangle ABC \sim \triangle AMP$ (AA); (ii) $\displaystyle \frac{CA}{PA} = \frac{BC}{MP}$.

Check: with $A(0, 0)$, $B(4, 0)$, $C(4, 3)$, $M(2, 0)$ and $P(2, 1.5)$, we get $\displaystyle \frac{CA}{PA} = \frac{5}{2.5} = 2$ and $\displaystyle \frac{BC}{MP} = \frac{3}{1.5} = 2$.

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) $\displaystyle \frac{CD}{GH} = \frac{AC}{FG}$ (ii) $\triangle DCB \sim \triangle HGE$ (iii) $\triangle DCA \sim \triangle HGF$.

What the given sim­i­lar­ity tells us. $\triangle ABC \sim \triangle FEG$, so $\angle A = \angle F$, $\angle B = \angle E$ and $\angle ACB = \angle FGE$. Halv­ing the last pair of equal angles gives

$$\displaystyle \angle ACD = \angle DCB = \tfrac12 \angle ACB = \tfrac12 \angle FGE = \angle FGH = \angle HGE.$$

(iii) In $\triangle DCA$ and $\triangle HGF$: $\angle DAC = \angle HFG$ (since $\angle A = \angle F$) and $\angle DCA = \angle HGF$ (shown above). By AA, $\triangle DCA \sim \triangle HGF$.

(ii) In $\triangle DCB$ and $\triangle HGE$: $\angle DBC = \angle HEG$ (since $\angle B = \angle E$) and $\angle DCB = \angle HGE$ (shown above). By AA, $\triangle DCB \sim \triangle HGE$.

(i) From (iii), with $D \leftrightarrow H$, $C \leftrightarrow G$, $A \leftrightarrow F$:

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

Answer: (i) $\displaystyle \frac{CD}{GH} = \frac{AC}{FG}$; (ii) $\triangle DCB \sim \triangle HGE$ (AA); (iii) $\triangle DCA \sim \triangle HGF$ (AA). Part (i) is eas­i­est to prove after part (iii), so we proved (iii) first.

Check: take $A(0, 0)$, $B(6, 0)$, $C(2, 4)$ and the half-size copy $F(10, 0)$, $E(13, 0)$, $G(11, 2)$. Locat­ing $D$ and $H$ with the angle bisec­tor the­o­rem gives $CD \approx 4.052$ and $GH \approx 2.026$. Their ratio is $2$, which equals $\displaystyle \frac{AC}{FG} = \frac{2\sqrt5}{\sqrt5}$. In (ii) and (iii), every pair of cor­re­spond­ing sides is also in the ratio $2$.

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 $\triangle ABD \sim \triangle ECF$.

Step 1. $AB = AC$, so the angles oppo­site these sides are equal: $\angle ABC = \angle ACB$.

Step 2. $D$ lies on $BC$, so $\angle ABD = \angle ABC$. $E$ lies on $CB$ pro­duced and $F$ lies on $CA$, so $\angle ECF = \angle BCA$. There­fore

$$\angle ABD = \angle ECF.$$

Step 3. $\angle ADB = 90^\circ$ (since $AD \perp BC$) and $\angle EFC = 90^\circ$ (since $EF \perp AC$), so $\angle ADB = \angle EFC$.

Step 4. By AA, $\triangle ABD \sim \triangle ECF$.

Answer: $\triangle ABD \sim \triangle ECF$ (AA), with $A \leftrightarrow E$, $B \leftrightarrow C$, $D \leftrightarrow F$.

Check with coor­di­nates: take $A(0, 4)$, $B(-3, 0)$, $C(3, 0)$, so $AB = AC = 5$ and $D = (0, 0)$. Take $E(-5, 0)$. The foot of the per­pen­dic­u­lar from $E$ to $AC$ is $F(0.12, 3.84)$. Then $\displaystyle \frac{AB}{EC} = \frac{5}{8}$, $\displaystyle \frac{BD}{CF} = \frac{3}{4.8}$ and $\displaystyle \frac{AD}{EF} = \frac{4}{6.4}$, and all three ratios equal $0.625$.

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 $\triangle ABC \sim \triangle PQR$.

Given:

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

Step 1. $D$ and $M$ are the mid­points of $BC$ and $QR$, so $\displaystyle BD = \tfrac12 BC$ and $\displaystyle QM = \tfrac12 QR$. There­fore

$$\displaystyle \frac{BD}{QM} = \frac{\tfrac12 BC}{\tfrac12 QR} = \frac{BC}{QR}.$$

Step 2. So in $\triangle ABD$ and $\triangle PQM$,

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

and by SSS, $\triangle ABD \sim \triangle PQM$. Hence $\angle ABD = \angle PQM$, which is the same as $\angle ABC = \angle PQR$.

Step 3. In $\triangle ABC$ and $\triangle PQR$, we have $\displaystyle \frac{AB}{PQ} = \frac{BC}{QR}$ and the included angles are equal, $\angle B = \angle Q$. By SAS,

$$\triangle ABC \sim \triangle PQR.$$

Answer: $\triangle ABC \sim \triangle PQR$ (SSS first, then SAS).

Check with num­bers: take $AB = 6$, $BC = 8$ (so $BD = 4$), $AD = 5$ and $PQ = 9$, $QR = 12$ (so $QM = 6$), $PM = 7.5$. Then $\displaystyle \frac69 = \frac46 = \frac{5}{7.5} = \frac23$.

Ques­tion 13#

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

Step 1. Com­pare $\triangle ADC$ and $\triangle BAC$.

• $\angle ADC = \angle BAC$ (given);
• $\angle ACD = \angle BCA$ (com­mon angle, since $D$ lies on $BC$).

By AA, $\triangle ADC \sim \triangle BAC$, with $A \leftrightarrow B$, $D \leftrightarrow A$, $C \leftrightarrow C$.

Step 2. The cor­re­spond­ing sides are pro­por­tional:

$$\displaystyle \frac{DC}{AC} = \frac{AC}{BC}.$$

Step 3. Cross-mul­ti­ply:

$$CA^2 = CB \cdot CD.$$

Answer: $CA^2 = CB \cdot CD$, as required.

Check: if $CB = 8$ and $CD = 2$, the result gives $CA^2 = 16$, so $CA = 4$. A coor­di­nate test with $A(1, 3)$, $B(-4, 0)$, $C(4, 0)$ and $D$ cho­sen so that $\displaystyle CD = \frac{CA^2}{CB}$ con­firms that $\angle ADC = \angle BAC$ exactly.

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 $\triangle ABC \sim \triangle PQR$.

Given: $\displaystyle \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.

Unlike Ques­tion 12, the side bisected by the median ($BC$) is not in the given ratio, so we need a con­struc­tion.

Con­struc­tion. Pro­duce $AD$ to $E$ so that $DE = AD$, and join $CE$ and $BE$. Sim­i­larly, pro­duce $PM$ to $N$ so that $MN = PM$, and join $RN$ and $QN$.

Step 1. In quadri­lat­eral $ABEC$, the diag­o­nals $AE$ and $BC$ bisect each other at $D$. So $ABEC$ is a par­al­lel­o­gram, and

$$CE = AB, \qquad BE = AC.$$

In the same way, $PQNR$ is a par­al­lel­o­gram, so $RN = PQ$ and $QN = PR$.

Step 2. Also $AE = 2AD$ and $PN = 2PM$. Using the given ratios,

$$\displaystyle \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM} \implies \frac{BE}{QN} = \frac{AB}{PQ} = \frac{AE}{PN}.$$

By SSS, $\triangle ABE \sim \triangle PQN$, so $\angle BAE = \angle QPN$ … (1)

Step 3. In the same way,

$$\displaystyle \frac{AC}{PR} = \frac{CE}{RN} = \frac{AE}{PN},$$

so by SSS, $\triangle ACE \sim \triangle PRN$, and $\angle CAE = \angle RPN$ … (2)

Step 4. Adding (1) and (2):

$$\angle BAE + \angle CAE = \angle QPN + \angle RPN \implies \angle BAC = \angle QPR.$$

Step 5. In $\triangle ABC$ and $\triangle PQR$, we have $\displaystyle \frac{AB}{PQ} = \frac{AC}{PR}$ and the included angles are equal, $\angle A = \angle P$. By SAS, $\triangle ABC \sim \triangle PQR$.

Answer: $\triangle ABC \sim \triangle PQR$.

Check with coor­di­nates: for $A(0, 0)$, $B(6, 0)$, $C(2, 4)$, the mid­point of $BC$ is $D(4, 2)$ and $E = (8, 4)$. $D$ is also the mid­point of $AE$. We get $CE = AB = 6$ and $BE = AC = 2\sqrt5 \approx 4.472$, exactly as the con­struc­tion requires.

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.

Step 1. Let the tow­er's height be $h$ m. The pole and the tower are both ver­ti­cal, so each makes a right angle with the ground. At the same moment, the sun's rays make the same angle with the ground in both cases. So by AA, the tri­an­gle formed by the pole and its shadow is sim­i­lar to the tri­an­gle formed by the tower and its shadow.

Step 2. In sim­i­lar tri­an­gles, cor­re­spond­ing sides are pro­por­tional:

$$\displaystyle \frac{\text{height of pole}}{\text{shadow of pole}} = \frac{\text{height of tower}}{\text{shadow of tower}} \implies \frac{6}{4} = \frac{h}{28}.$$

Step 3. Solve for $h$:

$$\displaystyle h = \frac{6 \times 28}{4} = \frac{168}{4} = 42.$$

Check: $\displaystyle \frac{6}{4} = 1.5$ and $\displaystyle \frac{42}{28} = 1.5$.

Answer: the tower is 42 m high.

Ques­tion 16#

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

The same result is often writ­ten as $\displaystyle \frac{AB}{AD} = \frac{PQ}{PM}$, which is this equa­tion rearranged. Both forms are proved below.

Step 1. $\triangle ABC \sim \triangle PQR$, so

$$\displaystyle \frac{AB}{PQ} = \frac{BC}{QR} \quad \text{and} \quad \angle B = \angle Q.$$

Step 2. $D$ and $M$ are mid­points, so

$$\displaystyle \frac{BD}{QM} = \frac{\tfrac12 BC}{\tfrac12 QR} = \frac{BC}{QR} = \frac{AB}{PQ}.$$

Step 3. In $\triangle ABD$ and $\triangle PQM$, we have $\displaystyle \frac{AB}{PQ} = \frac{BD}{QM}$ and the included angles are equal, $\angle ABD = \angle PQM$. By SAS, $\triangle ABD \sim \triangle PQM$.

Step 4. So

$$\displaystyle \frac{AB}{PQ} = \frac{AD}{PM}, \quad \text{and rearranging,} \quad \frac{AB}{AD} = \frac{PQ}{PM}.$$

Answer: $\displaystyle \frac{AB}{PQ} = \frac{AD}{PM}$ (equiv­a­lently, $\displaystyle \frac{AB}{AD} = \frac{PQ}{PM}$).

Check with num­bers: for $A(0, 0)$, $B(6, 0)$, $C(2, 4)$, we get $AB = 6$ and $AD = \sqrt{20} \approx 4.472$. A copy scaled by $\displaystyle \tfrac13$ has $PQ = 2$ and $PM \approx 1.491$. Then $\displaystyle \frac{AB}{AD} \approx 1.342$ and $\displaystyle \frac{PQ}{PM} \approx 1.342$.

Across the whole exer­cise, the same idea keeps com­ing back: find two equal angles for AA, or a ratio together with its included angle for SAS. Ques­tions 12 and 14 look alike, but Ques­tion 14 needs the dou­bled-median con­struc­tion, because the side cut in half by the median does not appear in its given ratio.

Key terms#

Sim­i­lar tri­an­gles

Tri­an­gles whose cor­re­spond­ing angles are equal and whose cor­re­spond­ing sides are in the same ratio.

Cor­re­spon­dence

The match­ing of ver­tices shown by the order of let­ters, for exam­ple $A \leftrightarrow P$, $B \leftrightarrow Q$, $C \leftrightarrow R$ in $\triangle ABC \sim \triangle PQR$.

AA cri­te­rion

Two pairs of equal angles are enough to prove that two tri­an­gles are sim­i­lar.

SSS cri­te­rion

All three pairs of cor­re­spond­ing sides are in the same ratio.

SAS cri­te­rion

One pair of equal angles, with the sides includ­ing those angles in the same ratio.

Included angle

The angle formed between two given sides of a tri­an­gle.

Median

The seg­ment join­ing a ver­tex of a tri­an­gle to the mid­point of the oppo­site side.

Alti­tude

The per­pen­dic­u­lar seg­ment from a ver­tex of a tri­an­gle to the line con­tain­ing the oppo­site side.

Com­mon ques­tions#

Can I write AAA instead of AA?#

Yes. NCERT names the cri­te­rion AAA. Two equal pairs of angles force the third pair to be equal, so AA is enough in prac­tice.

Why does the order of let­ters mat­ter?#

The order tells you which sides cor­re­spond. If you write $\triangle ABC \sim \triangle QRP$ when the cor­rect state­ment is $\triangle ABC \sim \triangle PQR$, the ratios you read off will be wrong.

When does SAS fail?#

SAS fails when the equal angle is not between the two pro­por­tional sides, as in the sec­ond illus­tra­tion in Ques­tion 1. Two sides and a non-included angle do not prove sim­i­lar­ity.

How do I spot a com­mon angle?#

Look for two tri­an­gles that share a ver­tex, where the sides from that ver­tex lie along the same two rays. This hap­pens in Ques­tions 5, 6, 7, 9 and 13.

Why is a con­struc­tion needed in Ques­tion 14?#

The given ratio does not include side $BC$, which the median bisects. Dou­bling the median cre­ates a par­al­lel­o­gram, which turns the median into a side of a tri­an­gle and makes SSS pos­si­ble.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Kise­lev, A. P. Kise­lev's Geom­e­try, Book I: Planime­try (adapted by A. Given­tal). Sum­iz­dat.