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

(i) △PQL ~ △RPM

(ii) QL × RM = PL × PM

(iii) PQ² = QR × QL

(i) From figure,

∠QRP = ∠MRP

and

∠LPQ = ∠MRP

Also,

∠RQP = ∠LQP

and

∠LQP = ∠RPM

In △PQL and △RPM,

∠LPQ = ∠MRP [Proved above]

∠LQP = ∠RPM [Proved above]

∴ △PQL ~ △RPM [By A.A. axiom]

Hence, proved that △PQL ~ △RPM.

(ii) We know that,

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \dfrac{QL}{PM} = \dfrac{PL}{RM}$

⇒ QL × RM = PL × PM

Hence, proved that QL × RM = PL × PM.

(iii) In △LPQ and △PQR,

∠Q = ∠Q [Common]

∠QPL = ∠PRQ [Given]

∴ △LPQ ~ △PQR [By A.A. axiom]

We know that,

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \dfrac{PQ}{QR} = \dfrac{QL}{PQ}$

⇒ PQ² = QL × QR

Hence, proved that PQ² = QL × QR.