Question

PQRS is a cyclic quadrilateral. If the bisector of angle QPS and PRS meet the circle at point A and B respectively. Prove that AB is the diameter of the circle.

Answer 1

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Answer 2

Answer:

∠AYB = 90°,  must be shows that AB is  diameter of the circle.

Step-by-step explanation:

Given the bisectors of angle W angle y of a circle quadrilateral.

WXYZ meet the circle at A and B resp.

A and B are joined

To prove that AB is two diameter of the circle Constructions, A and Y are joined

Sum of opposite angle of a cyclic quadrilateral being 180°

We have for cyclic quadrilateral WXYZ

∠XWZ + ∠XYZ = 180°

⇒$\frac{1}{2}$ ( ∠XWZ + ∠XYZ) =$\frac{1}{2}$(180)

⇒$\frac{1}{2}$∠XWZ +$\frac{1}{2}$∠XYZ=$\frac{1}{2}$(180)

⇒∠XWA + ∠XYB = 90°( since WA and YB are bisectors of ∠XWZ and ∠XYZ resp.)

But ∠XWA = ∠XYB , being the angles on same are AX

So we have,

∠XYA + ∠XYB = 90°

⇒∠AYB = 90°         ---------------(1)

Equation (1) must be shows that AB is two diameter of the circle.