Question

In a triangle PQR, L and M are two points on the base QR, such that $\angle LPQ=\angle QRP$ and $\angle RPM=\angle RQP$. Prove that
$P{Q}^2=QR\times QL$

Answer 1

In $△PQL$ and $△RMP$

$\angle LPQ=\angle QRP$(given)

$\angle RQP=\angle RPM$(given)

$\Rightarrow △PQL\sim △RMP$(AA similarity)

$\angle LPQ=\angle QRP$(given)

$\angle Q=\angle Q$(common)

$△PQL\sim △RQP$ by AA similarity

$=\frac{PQ}{RQ}=\frac{QL}{QP}=\frac{PL}{PR}$

$\Rightarrow {PQ}^2=QR\times QL$

Hence proved.