Question

PQR is an equilateral triangle. S is any point on PQ. Prove that PR > SR.

Answer 1

Given: ΔPQR is an equilateral triangle and S is any point on PQ.

Prove: PR > SR

As ΔPQR is equilateral triangle.

∴ ∠P = ∠Q = ∠R = 60°

In ΔSQR,

Exterior angle, ∠PSR = ∠SRQ + ∠SQR

∠PSR = ∠SRQ + ∠Q

∠PSR = ∠SRQ + ∠P   ...(As PQR is an equilateral triangle)

So, In ΔPSR, ∠PSR > ∠P

∴ PR > SR   ...(In any triangle, the side opposite to the larger angle is longer)

Hence, proved.