Question

If in a cyclic quadrilateral ABCD, opposite angle bisectors interects at points P and Q of circumcircle of this quadrilateral, then Prove that PQ is diameter of the circle

Answer 1

Given : ABCD is a cyclic quadrilateral, bisectors of ∠A and ∠C cuts the circle at Q and P

To Prove : PQ is diameter of this circle

Construction : Join PA and PD

Proof : To prove PQ is diameter of circle ∠QAP should be equal to 90°

Now, ABCD is a cyclic quadrilateral, then ∠A + ∠C = 180°

= \(\frac { 1 }{ 2 }\)∠A + \(\frac { 1 }{ 2 }\)∠C = 90°

= ∠QAD + ∠PCD = 90°

∠PCD and ∠PAD are the angles subtended by are PD is the same segment.

∴ ∠PCD = ∠PAD …(ii)

From (i) and (ii)

∠QAD + ∠PAD = 90° [∵ ∠QAD + ∠PAD = ∠QAP]

⇒ ∠QAP = 90°

⇒ ∠QAP is angle in semicircle.