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: [4 MARKS]
(i) $\Delta PQL\sim \Delta RPM$
(ii) $QL\times RM=PL\times PM$
(iii) $P{Q}^2=QR\times QL$

Answer 1

Application of theorems: 2 Marks
Calculation: 2 Marks



$\begin{array}{cccc}\left(i\right)& Proof& \angle LPQ=\angle QRP& \left[Given\right]\\ & and& \angle RPQ=\angle RPM& \left[Given\right]\\ & \therefore & \Delta PQL\sim \Delta RPM& \left[AAsimilarly\right]\\ \left(ii\right)& \because & \Delta PQL\sim \Delta RPM& \left(proved\right)\\ & \therefore & \frac{PQ}{RP}=\frac{QL}{PM}=\frac{PL}{RM}& \end{array}$
(Corresponding~sidesof~similar~triangles~are~proportional)
$\begin{array}{cccc}\left(iii\right)& & \angle LPQ=\angle QRP& \left[Given\right]\\ & & \angle LPQ=\angle QRP& \left[Given\right]\\ & & \angle Q=\angle Q& \left[Common\right]\\ & \therefore & \Delta PQL\sim \Delta RQP& \left[AAsimilarly\right]\\ & \Rightarrow & \frac{PQ}{RQ}=\frac{QL}{QP}=\frac{PL}{RP}& \\ & \Rightarrow & P{Q}^2=QR\times QL& Henceproved\end{array}$