NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.3 Triangles

1. $\Delta ABC$ and $\Delta DBC$ 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) $\Delta ABD \cong \Delta ACD$
(ii) $\Delta ABP \cong \Delta ACP$
(iii) $AP$ bisects $\angle A$ as well as $\angle D$.
(iv) $AP$ is the perpendicular bisector of $BC$.

(i) In $\Delta ABD$ and $\Delta ACD$:
1. $AB = AC$ (Given, $\Delta ABC$ is isosceles)
2. $BD = CD$ (Given, $\Delta DBC$ is isosceles)
3. $AD = AD$ (Common)
$\therefore$ $\Delta ABD \cong \Delta ACD$ (By SSS Rule).
$\Rightarrow \angle BAD = \angle CAD$ (CPCT) … (eq. 1)

(ii) In $\Delta ABP$ and $\Delta ACP$:
1. $AB = AC$ (Given)
2. $\angle BAP = \angle CAP$ (From eq. 1)
3. $AP = AP$ (Common)
$\therefore$ $\Delta ABP \cong \Delta ACP$ (By SAS Rule).
$\Rightarrow BP = CP$ and $\angle APB = \angle APC$ (CPCT) … (eq. 2)

(iii) AP bisects $\angle A$ from (eq. 1). Now for $\angle D$:
In $\Delta BDP$ and $\Delta CDP$:
1. $BD = CD$ (Given)
2. $BP = CP$ (From eq. 2)
3. $DP = DP$ (Common)
$\therefore \Delta BDP \cong \Delta CDP$ (By SSS Rule).
$\Rightarrow \angle BDP = \angle CDP$ (CPCT).
Hence, $AP$ bisects $\angle A$ as well as $\angle D$.

(iv) From (eq. 2), $\angle APB = \angle APC$.
Since $BC$ is a line, $\angle APB + \angle APC = 180^\circ$ (Linear Pair).
$2\angle APB = 180^\circ \Rightarrow \angle APB = 90^\circ$.
Also, $BP = CP$.
Therefore, $AP$ is the perpendicular bisector of $BC$.

2. $AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that:
(i) $AD$ bisects $BC$
(ii) $AD$ bisects $\angle A$.

Proof: In right triangles $\Delta ADB$ and $\Delta ADC$:
1. Hypotenuse $AB = AC$ (Given)
2. Side $AD = AD$ (Common)
3. $\angle ADB = \angle ADC = 90^\circ$ (Altitude)
$\therefore \Delta ADB \cong \Delta ADC$ (By RHS Rule).

(i) From CPCT:
$BD = CD$ (Hence, $AD$ bisects $BC$).

(ii) From CPCT:
$\angle BAD = \angle CAD$ (Hence, $AD$ bisects $\angle A$).

3. Two sides $AB$ and $BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\Delta PQR$. Show that:
(i) $\Delta ABM \cong \Delta PQN$
(ii) $\Delta ABC \cong \Delta PQR$

(i) Given $BC = QR$ and $AM, PN$ are medians.
So, $BM = \frac{1}{2}BC$ and $QN = \frac{1}{2}QR$. Since $BC = QR$, then $BM = QN$.
In $\Delta ABM$ and $\Delta PQN$:
1. $AB = PQ$ (Given)
2. $AM = PN$ (Given)
3. $BM = QN$ (Proved above)
$\therefore$ $\Delta ABM \cong \Delta PQN$ (By SSS Rule).
$\Rightarrow \angle B = \angle Q$ (CPCT) … (eq. 1)

(ii) In $\Delta ABC$ and $\Delta PQR$:
1. $AB = PQ$ (Given)
2. $\angle B = \angle Q$ (From eq. 1)
3. $BC = QR$ (Given)
$\therefore$ $\Delta ABC \cong \Delta PQR$ (By SAS Rule).

4. $BE$ and $CF$ are two equal altitudes of a triangle $ABC$. Using RHS congruence rule, prove that the triangle $ABC$ is isosceles.

Proof: Consider $\Delta BCF$ and $\Delta CBE$:
1. $\angle BFC = \angle CEB = 90^\circ$ (Altitudes)
2. Hypotenuse $BC = BC$ (Common)
3. Side $CF = BE$ (Given equal altitudes)
$\therefore \Delta BCF \cong \Delta CBE$ (By RHS Rule).

Conclusion:
From CPCT, $\angle FBC = \angle ECB$, which implies $\angle B = \angle C$.
Since angles opposite to equal sides are equal:
$AC = AB$. Hence, $\Delta ABC$ is isosceles.

5. $ABC$ is an isosceles triangle with $AB = AC$. Draw $AP \perp BC$ to show that $\angle B = \angle C$.

Proof: In right triangles $\Delta APB$ and $\Delta APC$:
1. Hypotenuse $AB = AC$ (Given)
2. Side $AP = AP$ (Common)
3. $\angle APB = \angle APC = 90^\circ$ (Construction)
$\therefore \Delta APB \cong \Delta APC$ (By RHS Rule).

Conclusion:
$\Rightarrow$ $\angle B = \angle C$ (By CPCT).