Question

14. S is a point on the side QR of a triangle PQR such that <PSR = <QPR. Show that
RP2 = RQ* RS​

Answer 1

Given

• ∠PSR = ∠QPR
• S is point on QR.

To Prove

RP² = RQ × RS

Proof

In ∆PQR and ∆PSR, we have

→ ∠QPR = ∠PSR (Given)

→ ∠R = ∠R (Common)

Hence, by AA - Criteria both triangles are similar.

Therefore by Thales theorem,

→ PR/RS = QR/PR

→ PR² = QR × RS

Or RP² = RQ * RS

Q.E.D

Answer 2

Step-by-step explanation:

In ΔPQR and ΔPSR ,

‹PSR = ‹QPR.... (Given)

So, by A.A. criteria (Angle - Angel)

ΔPSR ~ ΔPQR

By BPT (Basic Proportionality Theorem)

» PR/RS = QR/PR

___________[Cross multiply]

» PR² = QR * RS

____________________[Answer]