In an isosceles ∆ABC,

Question:

In an isosceles $\triangle \mathrm{ABC}$, the base $\mathrm{AB}$ is produced both the ways to $\mathrm{P}$ and $\mathrm{Q}$ such that $\mathrm{AP} \times \mathrm{BQ}=\mathrm{AC}^{2}$. Prove that $\triangle \mathrm{APC} \sim \triangle \mathrm{BCQ}$.

Solution:

It is given that $\triangle A B C$ is isosceles and $A P \times B Q=A C^{2}$.

We have to prove that $\triangle A P C \sim \triangle B C Q$.

It is given that $\triangle A B C$ is an isosceles triangle, so $\mathrm{AC}=\mathrm{BC}$.

Now,

$A P \times B Q=A C^{2} \quad$ (Given)

$A P \times B Q=A C \times A C$

$\Rightarrow \frac{A P}{A C}=\frac{A C}{B Q}$

$\Rightarrow \mathrm{APAC}=\mathrm{BCBQ}$

Also,

$\angle \mathrm{CAB}=\angle \mathrm{CBA}$ Equal sides have equal angles opposite to them $\Rightarrow 180^{\circ}-\angle \mathrm{CAP}=180^{\circ}-\angle \mathrm{CBQ} \Rightarrow \angle \mathrm{CAP}=\angle \mathrm{CBQ}$

Hence, $\triangle A P C \sim \triangle B C Q$ (SAS Similarity)