Question

In $\Delta PQR$, ray $QS$ bisects of $\angle PQR,P-S-R$. Show that $\frac{A\left(\Delta PQR\right)}{A\left(\Delta QRS\right)}=\frac{PQ}{QR}$

Answer 1

In $△PQR,$ ray $QS$ bisects $\angle PQR\therefore \angle PQS=\angle RQS$

By angle bisector theorem, $\frac{PQ}{PS}=\frac{QR}{RS}\Rightarrow \frac{PQ}{QR}=\frac{PS}{RS}.....\left(1\right)$

Let $h$ be a perpendicular drawn from vertex $Q$ to side $PR$

$\therefore$ Area of $△PQR=\frac{1}{2}\times PS\times h$ and Area of $△QRS=\frac{1}{2}\times RS\times h$

Thus, $\frac{ar\left(△PQR\right)}{ar\left(△QRS\right)}=\frac{\frac{1}{2}\times PS\times h}{\frac{1}{2}\times RS\times h}=\frac{PS}{RS}$

From equation (1)

$\frac{ar\left(△PQR\right)}{ar\left(△QRS\right)}=\frac{PQ}{QR}$