Question

A is any point in the angle PQR such that the perpendiculars drawn from A on PQ and QR are equal. Prove that ∠AQP = ∠AQR.

Answer 1

Given,
AM ⊥ PQ and AN ⊥ QR
AM = AN
In ΔAQM and ΔAQN,
AM = AN                   ....(given)
AQ = AQ                   ....(common)
∠AMQ - ∠ANQ       ....(Each = 90°)
So, by R.H.S. congruence, we have
ΔAQM ≅ ΔAQN
⇒ ∠AQM = ∠AQN  ....(c.p.c.t)
⇒ ∠AQP = ∠AQR.