Question

In a triangle PQR, L and M are two points on the base QR, such that ∠LPQ = ∠QRP and ∠RPM = ∠RQP. Prove that:

1. ΔPQL ∼ ΔRPM
2. QL × RM = PL × PM
3. PQ² = QR × QL

Answer 1

i. In ΔPQL and ΔRMP

∠LPQ = ∠QRP ...(Given)

∠RQP = ∠RPM  ...(Given)

ΔPQL ∼ ΔRMP  ...(AA similarity)

ii. As ΔPQL ∼ ΔRMP ...(Proved above)

${\displaystyle \frac{PQ}{RP}=\frac{QL}{PM}=\frac{PL}{RM}}$

${\displaystyle \Rightarrow }$ QL × RM = PL × PM

iii. ∠LPQ = ∠QRP ...(Given)

∠Q = ∠Q  ...(Common)

∆PQL ∼ ∆RQP  ...(AA similarity)

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

${\displaystyle \Rightarrow }$ PQ² = QR × QL