Exercise 7.3 Practice
SSS and RHS Congruence Criteria
Q1: Isosceles on Same Base
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$\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$. If $AD$ is extended to intersect $BC$ at $P$, show that:
(i) $\triangle ABD \cong \triangle ACD$
(ii) $\triangle ABP \cong \triangle ACP$
(iii) $AP$ bisects $\angle A$ as well as $\angle D$
(iv) $AP$ is the perpendicular bisector of $BC$
(i) $\triangle ABD \cong \triangle ACD$
(ii) $\triangle ABP \cong \triangle ACP$
(iii) $AP$ bisects $\angle A$ as well as $\angle D$
(iv) $AP$ is the perpendicular bisector of $BC$
(i) $\triangle ABD \cong \triangle ACD$:
1. $AB = AC$ (Given isosceles)
2. $BD = CD$ (Given isosceles)
3. $AD = AD$ (Common)
$\therefore \triangle ABD \cong \triangle ACD$ (SSS Rule).
1. $AB = AC$ (Given isosceles)
2. $BD = CD$ (Given isosceles)
3. $AD = AD$ (Common)
$\therefore \triangle ABD \cong \triangle ACD$ (SSS Rule).
(ii) $\triangle ABP \cong \triangle ACP$:
From (i), $\angle BAP = \angle CAP$ (CPCT).
1. $AB = AC$ (Given)
2. $\angle BAP = \angle CAP$ (Proved)
3. $AP = AP$ (Common)
$\therefore \triangle ABP \cong \triangle ACP$ (SAS Rule).
From (i), $\angle BAP = \angle CAP$ (CPCT).
1. $AB = AC$ (Given)
2. $\angle BAP = \angle CAP$ (Proved)
3. $AP = AP$ (Common)
$\therefore \triangle ABP \cong \triangle ACP$ (SAS Rule).
(iii) AP bisects $\angle A$ and $\angle D$:
From (i), $\angle BAD = \angle CAD$, so AP bisects $\angle A$.
In $\triangle BDP$ and $\triangle CDP$: $BD=CD$, $BP=CP$ (from ii), $DP=DP$.
$\triangle BDP \cong \triangle CDP$ (SSS).
$\Rightarrow \angle BDP = \angle CDP$. Thus AP bisects $\angle D$.
From (i), $\angle BAD = \angle CAD$, so AP bisects $\angle A$.
In $\triangle BDP$ and $\triangle CDP$: $BD=CD$, $BP=CP$ (from ii), $DP=DP$.
$\triangle BDP \cong \triangle CDP$ (SSS).
$\Rightarrow \angle BDP = \angle CDP$. Thus AP bisects $\angle D$.
(iv) AP is perpendicular bisector of BC:
From (ii), $BP = CP$.
Also $\angle APB = \angle APC$ (CPCT).
$\angle APB + \angle APC = 180^\circ$ (Linear Pair).
$2\angle APB = 180^\circ \Rightarrow \angle APB = 90^\circ$.
Thus, $AP \perp BC$ and bisects it.
From (ii), $BP = CP$.
Also $\angle APB = \angle APC$ (CPCT).
$\angle APB + \angle APC = 180^\circ$ (Linear Pair).
$2\angle APB = 180^\circ \Rightarrow \angle APB = 90^\circ$.
Thus, $AP \perp BC$ and bisects it.
Proved all parts.
Q2: Altitude of Isosceles
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$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that:
(i) $AD$ bisects $BC$
(ii) $AD$ bisects $\angle A$
(i) $AD$ bisects $BC$
(ii) $AD$ bisects $\angle A$
In right $\triangle ABD$ and right $\triangle ACD$:
1. $\angle ADB = \angle ADC = 90^\circ$ (Altitude)
2. $AB = AC$ (Hypotenuse, Given)
3. $AD = AD$ (Common Side)
1. $\angle ADB = \angle ADC = 90^\circ$ (Altitude)
2. $AB = AC$ (Hypotenuse, Given)
3. $AD = AD$ (Common Side)
$\therefore \triangle ABD \cong \triangle ACD$ (By RHS Rule).
(i) AD bisects BC:
By CPCT, $BD = CD$.
By CPCT, $BD = CD$.
(ii) AD bisects $\angle A$:
By CPCT, $\angle BAD = \angle CAD$.
By CPCT, $\angle BAD = \angle CAD$.
Proved.
Q3: Medians and Sides
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Two sides $AB$ and $BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\triangle PQR$. Show that:
(i) $\triangle ABM \cong \triangle PQN$
(ii) $\triangle ABC \cong \triangle PQR$
(i) $\triangle ABM \cong \triangle PQN$
(ii) $\triangle ABC \cong \triangle PQR$
Given $BC = QR$. Since $AM$ and $PN$ are medians:
$BM = \frac{1}{2}BC$ and $QN = \frac{1}{2}QR$.
$\therefore BM = QN$.
$BM = \frac{1}{2}BC$ and $QN = \frac{1}{2}QR$.
$\therefore BM = QN$.
(i) In $\triangle ABM$ and $\triangle PQN$:
1. $AB = PQ$ (Given)
2. $AM = PN$ (Given)
3. $BM = QN$ (Proved above)
$\therefore \triangle ABM \cong \triangle PQN$ (SSS Rule).
1. $AB = PQ$ (Given)
2. $AM = PN$ (Given)
3. $BM = QN$ (Proved above)
$\therefore \triangle ABM \cong \triangle PQN$ (SSS Rule).
(ii) In $\triangle ABC$ and $\triangle PQR$:
1. $AB = PQ$ (Given)
2. $\angle B = \angle Q$ (CPCT from part i)
3. $BC = QR$ (Given)
$\therefore \triangle ABC \cong \triangle PQR$ (SAS Rule).
1. $AB = PQ$ (Given)
2. $\angle B = \angle Q$ (CPCT from part i)
3. $BC = QR$ (Given)
$\therefore \triangle ABC \cong \triangle PQR$ (SAS Rule).
Proved.
Q4: Equal Altitudes (RHS)
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$BE$ and $CF$ are two equal altitudes of a triangle $ABC$. Using RHS congruence rule, prove that the triangle $ABC$ is isosceles.
In right $\triangle BEC$ and right $\triangle CFB$:
1. $\angle BEC = \angle CFB = 90^\circ$ (Altitudes)
2. $BC = CB$ (Common Hypotenuse)
3. $BE = CF$ (Given equal altitudes)
1. $\angle BEC = \angle CFB = 90^\circ$ (Altitudes)
2. $BC = CB$ (Common Hypotenuse)
3. $BE = CF$ (Given equal altitudes)
$\therefore \triangle BEC \cong \triangle CFB$ (By RHS Rule).
By CPCT, $\angle BCE = \angle CBF$.
i.e., $\angle C = \angle B$.
Sides opposite to equal angles are equal, so $AB = AC$.
Thus, $\triangle ABC$ is isosceles.
i.e., $\angle C = \angle B$.
Sides opposite to equal angles are equal, so $AB = AC$.
Thus, $\triangle ABC$ is isosceles.
Proved.
Q5: Perpendicular in Isosceles
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$ABC$ is an isosceles triangle with $AB = AC$. Draw $AP \perp BC$ to show that $\angle B = \angle C$.
In right $\triangle ABP$ and right $\triangle ACP$:
1. $\angle APB = \angle APC = 90^\circ$ (Given $AP \perp BC$)
2. $AB = AC$ (Given Hypotenuse)
3. $AP = AP$ (Common Side)
1. $\angle APB = \angle APC = 90^\circ$ (Given $AP \perp BC$)
2. $AB = AC$ (Given Hypotenuse)
3. $AP = AP$ (Common Side)
$\therefore \triangle ABP \cong \triangle ACP$ (By RHS Rule).
By CPCT, $\angle B = \angle C$.
Proved.