Question

In figure $\angle QPR={90}^{\circ },PS$ is its bisector.
If $ST\perp PR$, prove that $ST\times (PQ+PR)=PQ\times PR$.

Answer 1

In $△PQR$, since $PS$ is angle bisector & applying angle, bisector theorem
$\frac{PR}{PQ}=\frac{SR}{SQ}$
$△RTS\approx △RPQ$ (similarity)
$\frac{SR}{SQ}=\frac{TR}{TP}$
Given $\angle PTS={90}^{\circ }$
$\therefore$ In $\Delta PTS$, since $\angle TPS={45}^{\circ }$ (PS - angle
bisector)
$\angle PST$ also $={45}^{\circ }$
$\therefore \angle$ PTS is an isosceles $△$
$\Rightarrow PT=ST$
Using (2) in (1), we get $\frac{SR}{SQ}=\frac{TR}{ST}\dots ..\left(3\right)$
$TR=PR-PT$
$=PR-ST$
From $\left(A\right)$ and $\left(B\right)$, we get
$\begin{array}{cc}& \frac{PR}{PQ}=\frac{SR}{SQ}=\frac{TR}{ST}\\ & \therefore PR\times ST=TR\times PQ\\ & =(PR-ST)\times PQ\\ & =PR\times PQ-ST\times PQ\\ & \therefore PR\times ST+ST\times PQ=PR\times PQ\\ & \Rightarrow ST(PR+PQ)=PR\times PQ\end{array}$
Hence proved.