Question

If bisectors of opposite angles of a cyclic quadrilateral abcd intersect the circle, circumscribing it at the points p and q, prove that pq is a diameter of the circle.

Answer 1

as abcd is a cyclic quadrilateral then
b+d=180
1/2(b+d)=90 =angle abp+adq=90
and angle abp and PDA are in same segment so angle abp =pda
putting in eq so we get
angle PDA+adq =90
means angle PDQ=90
hence pq is diameter as angle in semi circle is 90
hence proved

Answer 2

Given: ABCD is a cyclic quadrilateral. AE and CF are the bisectors of ∠A and ∠C respectively.
 
To prove: EF is the diameter of the circle i.e. ∠EAF = 90°
 
Construction: Join AE and FD.
 
Proof: ABCD is a cyclic quadrilateral.
 
∴ ∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral are supplementary )
 
⇒ ∠EAD + ∠DCF = 90° ------- (1) (AE and CF are the bisector of ∠A and ∠C respectively)
 
∠DCF = ∠DAF -------- (2) (Angles in the same segment)
 
From equations (1) and (2), we get,
 
∠EAD + ∠DAF = 90°
 
⇒ ∠EAF = 90°
 
But ∠EAF is the angle in a semi-circle.
 
∴ EF is the diameter of the circle.