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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.
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Solution
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.
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