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

Applying theorems: 2 Marks
Calculation: 2 Marks

Let the trianglePQR be as shown in the given figure.

i) In $\Delta PQL$ and $\Delta RPM$,
$\angle LPQ=\angle QRP=\angle MRP$ [Given]
$\angle RQP=\angle LQP=\angle RPM$ [Given]
$\therefore \Delta PQL\sim \Delta RPM$ [By A.A. axiom of similarity]
Hence, proved.

ii) As $\Delta PQL\sim \Delta RPM$ [Proved in (i)]
$\therefore \frac{QL}{PM}=\frac{PL}{RM}$
$\Rightarrow QL\times RM=PL\times PM$
Hence, proved.

iii) In $\Delta LPQ$ and $\Delta PQR$
$\angle Q=\angle Q$ [Common]
$\angle QPL=\angle PRQ$ [Given]
$\therefore \Delta LPQ\sim \Delta PRQ$ [By A.A. axiom of similarity]
$\Rightarrow \frac{PQ}{QR}=\frac{QL}{PQ}$
$\Rightarrow P{Q}^2=QL\times QR$
Hence, proved.