Triangles - Exercise 6.3

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Overview

Master the three similarity criteria for triangles: Angle-Angle (AA/AAA), Side-Side-Side (SSS), and Side-Angle-Side (SAS). Apply criteria to solve angle puzzles, prove proportional sides, prove geometric theorems on altitudes and medians, and calculate real shadow heights.

NCERT Exercise Practice questions with detailed step-by-step solutions

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Criteria Drills (Q1-2) Proofs (Q3-6) Altitudes (Q7-9) Bisectors (Q10-11) Medians & Shadows (Q12-16)
Questions 1 & 2 Similarity Criteria Identification & Angles

1. State which pairs of triangles are similar. Write the similarity criterion used and the symbolic representation:
(i) $\Delta ABC$ with $\angle A = 40^\circ, \angle B = 60^\circ, \angle C = 80^\circ$ and $\Delta PQR$ with $\angle P = 40^\circ, \angle Q = 60^\circ, \angle R = 80^\circ$.
(ii) $\Delta DEF$ with sides $2, 2.5, 3$ and $\Delta LMN$ with sides $4, 5, 6$.

2. In the figure, $\Delta ODC \sim \Delta OBA$, $\angle BOC = 110^\circ$ and $\angle CDO = 60^\circ$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.

Question 1 Solution
(i) In $\Delta ABC$ and $\Delta PQR$:
$\angle A = \angle P = 40^\circ$, $\angle B = \angle Q = 60^\circ$, and $\angle C = \angle R = 80^\circ$.
Corresponding angles are equal.
$\therefore \Delta ABC \sim \Delta PQR$ by AAA (or AA) Similarity Criterion.

(ii) In $\Delta DEF$ and $\Delta LMN$:
$\dfrac{DE}{LM} = \dfrac{2}{4} = \dfrac{1}{2}$,   $\dfrac{EF}{MN} = \dfrac{2.5}{5} = \dfrac{1}{2}$,   $\dfrac{DF}{LN} = \dfrac{3}{6} = \dfrac{1}{2}$.
Since $\dfrac{DE}{LM} = \dfrac{EF}{MN} = \dfrac{DF}{LN} = \dfrac{1}{2}$, all three pairs of corresponding sides are proportional.
$\therefore \Delta DEF \sim \Delta LMN$ by SSS Similarity Criterion.
Question 2 Solution
1. Find $\angle DOC$:
$BD$ is a straight line, so $\angle DOC + \angle BOC = 180^\circ$ (Linear pair).
$\angle DOC = 180^\circ - 110^\circ = \mathbf{70^\circ}$.

2. Find $\angle DCO$:
In $\Delta ODC$, sum of angles is $180^\circ$:
$\angle DCO + \angle CDO + \angle DOC = 180^\circ$
$\angle DCO + 60^\circ + 70^\circ = 180^\circ \implies \angle DCO = 180^\circ - 130^\circ = \mathbf{50^\circ}$.

3. Find $\angle OAB$:
Since $\Delta ODC \sim \Delta OBA$, corresponding angles are equal:
$\angle OAB = \angle OCD = \angle DCO = \mathbf{50^\circ}$.
Questions 3, 4, 5 & 6 Proofs Using Similarity Criteria

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 $\dfrac{OA}{OC} = \dfrac{OB}{OD}$.

4. In the figure, $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\Delta PQS \sim \Delta TQR$.

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

6. In the figure, if $\Delta ABE \cong \Delta ACD$, show that $\Delta ADE \sim \Delta ABC$.

Question 3 Proof
In $\Delta OAB$ and $\Delta OCD$:
$\angle AOB = \angle COD$ (Vertically opposite angles)
$\angle OAB = \angle OCD$ (Alternate interior angles, as $AB \parallel DC$ and $AC$ is transversal)
$\angle OBA = \angle ODC$ (Alternate interior angles, as $AB \parallel DC$ and $BD$ is transversal)
$\therefore \Delta OAB \sim \Delta OCD$ by AAA (or AA) Similarity Criterion.
Since corresponding sides of similar triangles are proportional:
$\dfrac{OA}{OC} = \dfrac{OB}{OD}$. Hence proved.
Question 4 Proof
In $\Delta PQR$, given $\angle 1 = \angle 2 \implies PR = PQ$ (sides opposite to equal angles are equal).
Given ratio: $\dfrac{QR}{QS} = \dfrac{QT}{PR}$.
Substituting $PR = PQ$: $\dfrac{QR}{QS} = \dfrac{QT}{PQ} \implies \dfrac{PQ}{QT} = \dfrac{QS}{QR}$.
Now in $\Delta PQS$ and $\Delta TQR$:
1. $\dfrac{PQ}{QT} = \dfrac{QS}{QR}$
2. $\angle Q = \angle Q$ (Common angle)
$\therefore \Delta PQS \sim \Delta TQR$ by SAS Similarity Criterion. Hence proved.
Question 5 Proof
In $\Delta RPQ$ and $\Delta RTS$:
1. $\angle PRQ = \angle TRS$ (Common angle $\angle R$)
2. $\angle RPQ = \angle RTS$ (Given)
$\therefore \Delta RPQ \sim \Delta RTS$ by AA Similarity Criterion. Hence proved.
Question 6 Proof
Given $\Delta ABE \cong \Delta ACD$. By CPCT:
$AB = AC$    (1)
$AE = AD \implies AD = AE$    (2)
Dividing (2) by (1): $\dfrac{AD}{AB} = \dfrac{AE}{AC}$.
Now in $\Delta ADE$ and $\Delta ABC$:
1. $\dfrac{AD}{AB} = \dfrac{AE}{AC}$
2. $\angle DAE = \angle BAC$ (Common angle $\angle A$)
$\therefore \Delta ADE \sim \Delta ABC$ by SAS Similarity Criterion. Hence proved.
Questions 7, 8 & 9 Altitudes & Right Triangle Similarities

7. In the figure, altitudes $AD$ and $CE$ of $\Delta ABC$ intersect at $P$. Show that:
(i) $\Delta AEP \sim \Delta CDP$
(ii) $\Delta ABD \sim \Delta CBE$
(iii) $\Delta AEP \sim \Delta ADB$
(iv) $\Delta PDC \sim \Delta BEC$

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

9. In the figure, $ABC$ and $AMP$ are two right triangles, right-angled at $B$ and $M$ respectively. Prove that:
(i) $\Delta ABC \sim \Delta AMP$
(ii) $\dfrac{CA}{PA} = \dfrac{BC}{MP}$

Question 7 Solutions
(i) In $\Delta AEP$ and $\Delta CDP$:
$\angle AEP = \angle CDP = 90^\circ$ (Given altitudes)
$\angle APE = \angle CPD$ (Vertically opposite angles)
$\therefore \Delta AEP \sim \Delta CDP$ by AA criterion.

(ii) In $\Delta ABD$ and $\Delta CBE$:
$\angle ADB = \angle CEB = 90^\circ$
$\angle ABD = \angle CBE$ (Common angle $\angle B$)
$\therefore \Delta ABD \sim \Delta CBE$ by AA criterion.

(iii) In $\Delta AEP$ and $\Delta ADB$:
$\angle AEP = \angle ADB = 90^\circ$
$\angle PAE = \angle DAB$ (Common angle $\angle A$)
$\therefore \Delta AEP \sim \Delta ADB$ by AA criterion.

(iv) In $\Delta PDC$ and $\Delta BEC$:
$\angle PDC = \angle BEC = 90^\circ$
$\angle PCD = \angle BCE$ (Common angle $\angle C$)
$\therefore \Delta PDC \sim \Delta BEC$ by AA criterion.
Question 8 Solution
In $\Delta ABE$ and $\Delta CFB$:
1. $\angle A = \angle C$ (Opposite angles of parallelogram $ABCD$ are equal)
2. Since $AE \parallel BC$ (as $AD \parallel BC$ extended) and $BE$ is transversal:
$\angle AEB = \angle CBF$ (Alternate interior angles)
$\therefore \Delta ABE \sim \Delta CFB$ by AA Similarity Criterion. Hence proved.
Question 9 Solution
(i) In $\Delta ABC$ and $\Delta AMP$:
$\angle ABC = \angle AMP = 90^\circ$ (Given right angles)
$\angle BAC = \angle MAP$ (Common angle $\angle A$)
$\therefore \Delta ABC \sim \Delta AMP$ by AA criterion. Hence proved.

(ii) Since $\Delta ABC \sim \Delta AMP$, corresponding sides are proportional:
$\dfrac{CA}{PA} = \dfrac{BC}{MP} = \dfrac{AB}{AM} \implies \dfrac{CA}{PA} = \dfrac{BC}{MP}$. Hence proved.
Questions 10 & 11 Angle Bisectors & Isosceles Altitudes

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 $\Delta ABC$ and $\Delta EFG$. If $\Delta ABC \sim \Delta FEG$, show that:
(i) $\dfrac{CD}{GH} = \dfrac{AC}{FG}$
(ii) $\Delta DCB \sim \Delta HGE$
(iii) $\Delta DCA \sim \Delta HGF$

11. In the figure, $E$ 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 $\Delta ABD \sim \Delta ECF$.

Question 10 Solution
Given $\Delta ABC \sim \Delta FEG \implies \angle A = \angle F, \ \angle B = \angle E, \ \angle ACB = \angle EGF$.
Since $CD$ bisects $\angle ACB$ and $GH$ bisects $\angle EGF$:
$\angle ACD = \angle DCB = \dfrac{1}{2}\angle ACB$ and $\angle FGH = \angle HGE = \dfrac{1}{2}\angle EGF$.
Therefore: $\angle ACD = \angle FGH$ and $\angle DCB = \angle HGE$.

(iii) Proof of $\Delta DCA \sim \Delta HGF$:
In $\Delta DCA$ and $\Delta HGF$:
$\angle A = \angle F$ (from similarity of parent triangles)
$\angle ACD = \angle FGH$ (half of equal angles)
$\therefore \Delta DCA \sim \Delta HGF$ by AA criterion. Hence proved (iii).

(i) Proof of $\dfrac{CD}{GH} = \dfrac{AC}{FG}$:
Since $\Delta DCA \sim \Delta HGF$, their corresponding sides are proportional:
$\dfrac{CD}{GH} = \dfrac{AC}{FG}$. Hence proved (i).

(ii) Proof of $\Delta DCB \sim \Delta HGE$:
In $\Delta DCB$ and $\Delta HGE$:
$\angle B = \angle E$ (from similarity of parent triangles)
$\angle DCB = \angle HGE$ (half of equal angles)
$\therefore \Delta DCB \sim \Delta HGE$ by AA criterion. Hence proved (ii).
Question 11 Solution
In isosceles $\Delta ABC$, $AB = AC \implies \angle B = \angle C$ (angles opposite to equal sides are equal).
Now in $\Delta ABD$ and $\Delta ECF$:
1. $\angle ADB = \angle EFC = 90^\circ$ (Since $AD \perp BC$ and $EF \perp AC$)
2. $\angle ABD = \angle ECF$ (Since $\angle B = \angle C$ and $E-B-C$ is a straight line)
$\therefore \Delta ABD \sim \Delta ECF$ by AA Similarity Criterion. Hence proved.
Questions 12 to 16 Medians, Proportionality & Real Shadows

12. Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\Delta PQR$. Show that $\Delta ABC \sim \Delta PQR$.

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

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

15. A vertical pole of length $5\text{ m}$ casts a shadow $3\text{ m}$ long. At the same time, a tower casts a shadow $24\text{ m}$ long. Find the height of the tower.

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

Question 12 Solution
Given $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$.
Since $AD$ and $PM$ are medians, $D$ is mid-point of $BC \implies BC = 2BD$, and $M$ is mid-point of $QR \implies QR = 2QM$.
$\dfrac{AB}{PQ} = \dfrac{2BD}{2QM} = \dfrac{AD}{PM} \implies \dfrac{AB}{PQ} = \dfrac{BD}{QM} = \dfrac{AD}{PM}$.
$\therefore \Delta ABD \sim \Delta PQM$ by SSS similarity criterion.
This gives $\angle B = \angle Q$ (corresponding angles).
Now in $\Delta ABC$ and $\Delta PQR$:
1. $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ (Given)
2. $\angle B = \angle Q$ (Proved above)
$\therefore \Delta ABC \sim \Delta PQR$ by SAS Similarity Criterion. Hence proved.
Question 13 Solution
In $\Delta ABC$ and $\Delta DAC$:
1. $\angle BAC = \angle ADC$ (Given)
2. $\angle BCA = \angle ACD$ (Common angle $\angle C$)
$\therefore \Delta ABC \sim \Delta DAC$ by AA Similarity Criterion.
Corresponding sides are in the same ratio:
$\dfrac{CB}{CA} = \dfrac{CA}{CD} \implies CA \times CA = CB \times CD \implies CA^2 = CB \cdot CD$. Hence proved.
Question 14 Solution
Given: $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$.
Construction: Produce $AD$ to $E$ such that $AD = DE$. Join $CE$. Similarly, produce $PM$ to $N$ such that $PM = MN$ and join $RN$.
1. In $\Delta ABD$ and $\Delta ECD$, $BD = CD$ (median), $AD = DE$, $\angle ADB = \angle EDC$ (vertically opposite).
$\therefore \Delta ABD \cong \Delta ECD \implies AB = CE$. Similarly, $PQ = RN$.
2. Now $\dfrac{CE}{RN} = \dfrac{AC}{PR} = \dfrac{2AD}{2PM} = \dfrac{AE}{PN}$.
$\therefore \Delta ACE \sim \Delta PRN$ by SSS criterion $\implies \angle CAD = \angle RPM$.
3. Similarly, by joining $BE$ and $QN$, $\angle BAD = \angle QPM$.
Adding: $\angle CAD + \angle BAD = \angle RPM + \angle QPM \implies \angle A = \angle P$.
4. With $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$ and $\angle A = \angle P$, $\Delta ABC \sim \Delta PQR$ by SAS Criterion. Hence proved.
Question 15 Solution
Let the pole be $AB = 5\text{ m}$ with shadow $BC = 3\text{ m}$.
Let the tower be $PQ = h\text{ m}$ with shadow $QR = 24\text{ m}$.
At the same time, the sun's altitude (angle of elevation) is the same: $\angle C = \angle R$.
Both the pole and the tower are vertical to the ground: $\angle B = \angle Q = 90^\circ$.
$\therefore \Delta ABC \sim \Delta PQR$ by AA similarity criterion.
$\dfrac{AB}{PQ} = \dfrac{BC}{QR} \implies \dfrac{5}{h} = \dfrac{3}{24} \implies \dfrac{5}{h} = \dfrac{1}{8} \implies h = 5 \times 8 = 40\text{ m}$.
$\therefore$ The height of the tower is $40\text{ metres}$.
Question 16 Solution
Given $\Delta ABC \sim \Delta PQR \implies \angle B = \angle Q$ and $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$.
Since $AD$ and $PM$ are medians, $BC = 2BD$ and $QR = 2QM$.
$\dfrac{BC}{QR} = \dfrac{2BD}{2QM} = \dfrac{BD}{QM}$.
Therefore, $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$.
In $\Delta ABD$ and $\Delta PQM$:
1. $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$
2. $\angle B = \angle Q$
$\therefore \Delta ABD \sim \Delta PQM$ by SAS Similarity Criterion.
Hence, their corresponding sides are proportional: $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$. Hence proved.
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