Exercise 6.3 Practice
Master Similarity Criteria (AAA, SSS, SAS)
Q1: Identify Similarity
State which pairs of triangles are similar. Write the similarity criterion used and the symbolic representation.
(i) Triangles ABC and PQR
Analysis:
$\angle A = \angle P = 40^\circ$
$\angle B = \angle Q = 60^\circ$
$\angle C = \angle R = 80^\circ$
$\angle A = \angle P = 40^\circ$
$\angle B = \angle Q = 60^\circ$
$\angle C = \angle R = 80^\circ$
$\Delta ABC \sim \Delta PQR$ by AAA Criterion.
(ii) Triangles DEF and LMN
Analysis of Ratios:
$\frac{DE}{LM} = \frac{2}{4} = \frac{1}{2}$
$\frac{EF}{MN} = \frac{2.5}{5} = \frac{1}{2}$
$\frac{DF}{LN} = \frac{3}{6} = \frac{1}{2}$
$\frac{DE}{LM} = \frac{2}{4} = \frac{1}{2}$
$\frac{EF}{MN} = \frac{2.5}{5} = \frac{1}{2}$
$\frac{DF}{LN} = \frac{3}{6} = \frac{1}{2}$
$\Delta DEF \sim \Delta LMN$ by SSS Criterion.
Q2: Finding Angles
In the figure, $\Delta OD C \sim \Delta OBA$, $\angle BOC = 110^\circ$ and $\angle CDO = 60^\circ$. Find $\angle DOC$, $\angle DCO$ and $\angle OAB$.
1. Find $\angle DOC$:
$\angle DOC + \angle BOC = 180^\circ$ (Linear Pair)
$\angle DOC = 180 - 110 = 70^\circ$.
$\angle DOC + \angle BOC = 180^\circ$ (Linear Pair)
$\angle DOC = 180 - 110 = 70^\circ$.
2. Find $\angle DCO$:
In $\Delta ODC$: $\angle D + \angle DOC + \angle C = 180^\circ$
$60 + 70 + \angle C = 180$
$\angle DCO = 180 - 130 = 50^\circ$.
In $\Delta ODC$: $\angle D + \angle DOC + \angle C = 180^\circ$
$60 + 70 + \angle C = 180$
$\angle DCO = 180 - 130 = 50^\circ$.
3. Find $\angle OAB$:
Since $\Delta ODC \sim \Delta OBA$:
$\angle OAB = \angle OCD$ (Corresponding angles)
$\angle OAB = 50^\circ$.
Since $\Delta ODC \sim \Delta OBA$:
$\angle OAB = \angle OCD$ (Corresponding angles)
$\angle OAB = 50^\circ$.
$\angle DOC = 70^\circ, \angle DCO = 50^\circ, \angle OAB = 50^\circ$.
Q3: Trapezium Similarity
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that $\frac{OA}{OC} = \frac{OB}{OD}$.
Step 1: Identify Triangles
Consider $\Delta OAB$ and $\Delta OCD$.
Consider $\Delta OAB$ and $\Delta OCD$.
Step 2: Compare Angles
1. $\angle AOB = \angle COD$ (Vertically Opposite)
2. $\angle OAB = \angle OCD$ (Alt. Interior angles, AB || DC)
1. $\angle AOB = \angle COD$ (Vertically Opposite)
2. $\angle OAB = \angle OCD$ (Alt. Interior angles, AB || DC)
Step 3: Similarity
By AA Criterion, $\Delta OAB \sim \Delta OCD$.
By AA Criterion, $\Delta OAB \sim \Delta OCD$.
Step 4: Ratios
$\frac{OA}{OC} = \frac{OB}{OD}$ (Corresponding sides of similar triangles).
$\frac{OA}{OC} = \frac{OB}{OD}$ (Corresponding sides of similar triangles).
Hence Proved.
Q4: SAS Similarity Proof
In the figure, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\Delta PQS \sim \Delta TQR$.
Step 1: Isosceles Property
In $\Delta PQR$, given $\angle 1 = \angle 2$ (at Q and R).
Therefore, $PQ = PR$ (Sides opposite equal angles).
In $\Delta PQR$, given $\angle 1 = \angle 2$ (at Q and R).
Therefore, $PQ = PR$ (Sides opposite equal angles).
Step 2: Substitute
Given $\frac{QR}{QS} = \frac{QT}{PR}$.
Replace $PR$ with $PQ$: $\frac{QR}{QS} = \frac{QT}{PQ}$.
Given $\frac{QR}{QS} = \frac{QT}{PR}$.
Replace $PR$ with $PQ$: $\frac{QR}{QS} = \frac{QT}{PQ}$.
Step 3: Similarity
In $\Delta PQS$ and $\Delta TQR$:
1. $\frac{QS}{QR} = \frac{PQ}{QT}$ (Inverted from Step 2)
2. $\angle Q = \angle Q$ (Common)
By SAS Criterion: $\Delta PQS \sim \Delta TQR$.
In $\Delta PQS$ and $\Delta TQR$:
1. $\frac{QS}{QR} = \frac{PQ}{QT}$ (Inverted from Step 2)
2. $\angle Q = \angle Q$ (Common)
By SAS Criterion: $\Delta PQS \sim \Delta TQR$.
Hence Proved.
Q5: Points on Sides
S and T are points on sides PR and QR of $\Delta PQR$ such that $\angle P = \angle RTS$. Show that $\Delta RPQ \sim \Delta RTS$.
Proof:
In $\Delta RPQ$ and $\Delta RTS$:
1. $\angle R = \angle R$ (Common Angle)
2. $\angle P = \angle RTS$ (Given)
1. $\angle R = \angle R$ (Common Angle)
2. $\angle P = \angle RTS$ (Given)
$\Delta RPQ \sim \Delta RTS$ (AA Criterion).
Q6: Congruence to Similarity
In the figure, if $\Delta ABE \cong \Delta ACD$, show that $\Delta ADE \sim \Delta ABC$.
Step 1: Congruence
Given $\Delta ABE \cong \Delta ACD$.
So, $AB = AC$ and $AE = AD$ (CPCT).
Given $\Delta ABE \cong \Delta ACD$.
So, $AB = AC$ and $AE = AD$ (CPCT).
Step 2: Proportionality
Therefore, $\frac{AD}{AB} = \frac{AE}{AC}$.
Therefore, $\frac{AD}{AB} = \frac{AE}{AC}$.
Step 3: Similarity
In $\Delta ADE$ and $\Delta ABC$:
1. $\frac{AD}{AB} = \frac{AE}{AC}$
2. $\angle A = \angle A$ (Common)
Thus, $\Delta ADE \sim \Delta ABC$ (SAS Criterion).
In $\Delta ADE$ and $\Delta ABC$:
1. $\frac{AD}{AB} = \frac{AE}{AC}$
2. $\angle A = \angle A$ (Common)
Thus, $\Delta ADE \sim \Delta ABC$ (SAS Criterion).
Q7: Altitudes & Similarity
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) $\Delta PDC \sim \Delta BEC$
(i) $\Delta AEP \sim \Delta CDP$
(ii) $\Delta ABD \sim \Delta CBE$
(iii) $\Delta AEP \sim \Delta ADB$
(iv) $\Delta PDC \sim \Delta BEC$
(i) $\Delta AEP \sim \Delta CDP$:
$\angle AEP = \angle CDP = 90^\circ$
$\angle APE = \angle CPD$ (Vertically Opp)
$\therefore$ AA Similarity.
$\angle AEP = \angle CDP = 90^\circ$
$\angle APE = \angle CPD$ (Vertically Opp)
$\therefore$ AA Similarity.
(ii) $\Delta ABD \sim \Delta CBE$:
$\angle ADB = \angle CEB = 90^\circ$
$\angle B = \angle B$ (Common)
$\therefore$ AA Similarity.
$\angle ADB = \angle CEB = 90^\circ$
$\angle B = \angle B$ (Common)
$\therefore$ AA Similarity.
(iii) $\Delta AEP \sim \Delta ADB$:
$\angle AEP = \angle ADB = 90^\circ$
$\angle A = \angle A$ (Common)
$\therefore$ AA Similarity.
$\angle AEP = \angle ADB = 90^\circ$
$\angle A = \angle A$ (Common)
$\therefore$ AA Similarity.
(iv) $\Delta PDC \sim \Delta BEC$:
$\angle PDC = \angle BEC = 90^\circ$
$\angle C = \angle C$ (Common)
$\therefore$ AA Similarity.
$\angle PDC = \angle BEC = 90^\circ$
$\angle C = \angle C$ (Common)
$\therefore$ AA Similarity.
Q8: Parallelogram Extension
E is a point on side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\Delta ABE \sim \Delta CFB$.
1. In parallelogram ABCD, $\angle A = \angle C$ (Opposite angles).
2. AE || BC (Since AD || BC).
So, $\angle AEB = \angle CBF$ (Alternate Interior Angles).
So, $\angle AEB = \angle CBF$ (Alternate Interior Angles).
3. In $\Delta ABE$ and $\Delta CFB$:
i) $\angle A = \angle C$
ii) $\angle AEB = \angle CBF$
$\therefore \Delta ABE \sim \Delta CFB$ (AA Criterion).
i) $\angle A = \angle C$
ii) $\angle AEB = \angle CBF$
$\therefore \Delta ABE \sim \Delta CFB$ (AA Criterion).
Q9: Right Triangles
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) $\frac{CA}{PA} = \frac{BC}{MP}$
(i) $\Delta ABC \sim \Delta AMP$
(ii) $\frac{CA}{PA} = \frac{BC}{MP}$
(i) Proof:
In $\Delta ABC$ and $\Delta AMP$:
1. $\angle B = \angle M = 90^\circ$ (Given)
2. $\angle A = \angle A$ (Common)
$\therefore \Delta ABC \sim \Delta AMP$ (AA Criterion)
In $\Delta ABC$ and $\Delta AMP$:
1. $\angle B = \angle M = 90^\circ$ (Given)
2. $\angle A = \angle A$ (Common)
$\therefore \Delta ABC \sim \Delta AMP$ (AA Criterion)
(ii) Proof:
Since triangles are similar, corresponding sides are proportional:
$\frac{CA}{PA} = \frac{BC}{MP}$.
Since triangles are similar, corresponding sides are proportional:
$\frac{CA}{PA} = \frac{BC}{MP}$.
Q10: Angle Bisectors
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) $\frac{CD}{GH} = \frac{AC}{FG}$
(ii) $\Delta DCB \sim \Delta HGE$
(iii) $\Delta DCA \sim \Delta HGF$
(i) $\frac{CD}{GH} = \frac{AC}{FG}$
(ii) $\Delta DCB \sim \Delta HGE$
(iii) $\Delta DCA \sim \Delta HGF$
Given $\Delta ABC \sim \Delta FEG$, so $\angle C = \angle G$.
Halves are equal: $\angle ACD = \angle FGH$ and $\angle DCB = \angle HGE$.
Halves are equal: $\angle ACD = \angle FGH$ and $\angle DCB = \angle HGE$.
(i) & (iii) In $\Delta DCA$ and $\Delta HGF$:
$\angle A = \angle F$ (Given)
$\angle ACD = \angle FGH$ (Proved)
$\therefore \Delta DCA \sim \Delta HGF$ (AA).
$\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}$.
$\angle A = \angle F$ (Given)
$\angle ACD = \angle FGH$ (Proved)
$\therefore \Delta DCA \sim \Delta HGF$ (AA).
$\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}$.
(ii) Similarly for $\Delta DCB$ and $\Delta HGE$ using $\angle B = \angle E$.
Q11: Isosceles Triangle
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 $\Delta ABD \sim \Delta ECF$.
1. $AB = AC \Rightarrow \angle B = \angle C$.
2. In $\Delta ABD$ and $\Delta ECF$:
i) $\angle ABD = \angle ECF$ ($\angle B = \angle C$)
ii) $\angle ADB = \angle EFC = 90^\circ$ (Given)
i) $\angle ABD = \angle ECF$ ($\angle B = \angle C$)
ii) $\angle ADB = \angle EFC = 90^\circ$ (Given)
$\Delta ABD \sim \Delta ECF$ (AA Criterion).
Q12: Sides and Median
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 $\Delta ABC \sim \Delta PQR$.
Given $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.
Since medians bisect base, $BC=2BD, QR=2QM$.
$\Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.
$\Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.
$\Delta ABD \sim \Delta PQM$ (SSS). Thus $\angle B = \angle Q$.
In $\Delta ABC$ and $\Delta PQR$:
$\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
$\therefore \Delta ABC \sim \Delta PQR$ (SAS).
$\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
$\therefore \Delta ABC \sim \Delta PQR$ (SAS).
Q13: Geometric Mean
D is a point on the side BC of a triangle ABC such that $\angle ADC = \angle BAC$. Show that $CA^2 = CB \cdot CD$.
In $\Delta ABC$ and $\Delta DAC$:
1. $\angle BAC = \angle ADC$ (Given)
2. $\angle C = \angle C$ (Common)
1. $\angle BAC = \angle ADC$ (Given)
2. $\angle C = \angle C$ (Common)
$\therefore \Delta ABC \sim \Delta DAC$ (AA).
Ratio: $\frac{CA}{CD} = \frac{CB}{CA} \Rightarrow CA^2 = CB \cdot CD$.
Q14: Proportional Sides (Hard)
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 $\Delta ABC \sim \Delta PQR$.
Construction: Extend AD to E such that AD=DE. Join BE, CE.
1. Prove $\Delta ABD \cong \Delta ECD \Rightarrow AB=CE$.
2. Use proportionality to show $\Delta ACE \sim \Delta PRN$ (where N is ext point for PQR).
3. This gives $\angle CAD = \angle RPM$. Similarly $\angle BAD = \angle QPM$.
4. Total $\angle A = \angle P$. With proportional sides, $\Delta ABC \sim \Delta PQR$ (SAS).
Q15: Shadows
A vertical pole of length 5 m casts a shadow 3 m long. At the same time, a tower casts a shadow 24 m long. Find the height of the tower.
Similar triangles by AA (90° and sun's angle).
$\frac{h}{5} = \frac{24}{3}$
$h = \frac{24}{3} \times 5 = 8 \times 5 = 40$ m.
Height = 40 m
Q16: Median Ratio Proof
If AD and PM are medians of $\Delta ABC$ and $\Delta PQR$ respectively, where $\Delta ABC \sim \Delta PQR$, prove that $\frac{AB}{PQ} = \frac{AD}{PM}$.
1. $\Delta ABC \sim \Delta PQR \Rightarrow \angle B = \angle Q$ and $\frac{AB}{PQ} = \frac{BC}{QR}$.
2. Medians bisect BC and QR, so $\frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM}$.
3. In $\Delta ABD$ and $\Delta PQM$:
$\frac{AB}{PQ} = \frac{BD}{QM}$ and $\angle B = \angle Q$.
$\therefore \Delta ABD \sim \Delta PQM$ (SAS).
$\frac{AB}{PQ} = \frac{BD}{QM}$ and $\angle B = \angle Q$.
$\therefore \Delta ABD \sim \Delta PQM$ (SAS).
Hence $\frac{AB}{PQ} = \frac{AD}{PM}$.