Question

$$\begin{gathered} \mathrm{In}△\mathrm{PQR},\mathrm{if}\mathrm{QS}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle \mathrm{Q},\mathrm{then}\mathrm{show}\mathrm{that} \\ \frac{\mathrm{A}\left(△\mathrm{PQR}\right)}{\mathrm{A}\left(△\mathrm{QRS}\right)}=\frac{\mathrm{PQ}}{\mathrm{QR}}. \end{gathered}$$

Answer 1

Disclaimer :
$\mathrm{Rather}\mathrm{than}\frac{\mathrm{A}\left(△\mathrm{PQR}\right)}{\mathrm{A}\left(△\mathrm{QRS}\right)}=\frac{\mathrm{PQ}}{\mathrm{QR}}\mathrm{it}\mathrm{should}\mathrm{be}\frac{\mathrm{A}\left(△\mathrm{PQR}\right)}{\mathrm{A}\left(△\mathrm{QRS}\right)}=\frac{\mathrm{PR}}{\mathrm{SR}}$

$\mathrm{Construction}:\mathrm{Draw}\mathrm{QT}\perp \mathrm{PR}$.

$$\begin{gathered} \mathrm{Proof}: \\ \frac{\mathrm{A}\left(△\mathrm{PQR}\right)}{\mathrm{A}\left(△\mathrm{QRS}\right)}=\frac{{\displaystyle \frac{1}{2}}\times \mathrm{PR}\times \mathrm{TQ}}{{\displaystyle \frac{1}{2}}\times \mathrm{SR}\times \mathrm{TQ}} \\ =\frac{\mathrm{PR}}{\mathrm{SR}} \end{gathered}$$

Hence proved.