Question

In a $△PQR,PQ=PR,X$ is a point on $PR$ such that $Q{R}^2=PR\times XR$. Prove that $QX=QR$.

Answer 1

In ${\Delta }^u$ $PQR$ & $Q\times R_1$

$\frac{PR}{QR}=\frac{QR}{XR}$ $(\because {QR}^2=PR\times R)$

$\measuredangle R=\measuredangle R$

$\therefore$ $\Delta PQR\sim \Delta Q\times R$ (SAS criterion)

$\therefore$ $\frac{PQ}{QX}=\frac{QR}{XR}=\frac{PR}{QR}$ (corresponding parts)

$\Rightarrow \frac{PQ}{QX}=\frac{QR}{VR}$

or, $\frac{PR}{QX}=\frac{QR}{XR}$ $(\because PQ=PR)$

$\Rightarrow QR=QX$ proved.