In the given figure, ABCD is a cyclic quadrilateral. A circle passing through A and B meets AD and BC in the points E and F respectively. Prove that EF || DC.

Question

Philoid · Accepted Answer

Here,

In cyclic Quadrilateral ABFE

∠ABF + ∠AEF = 180° (opposite angles in cyclic quadrilateral are supplementary) –1

In cyclic Quadrilateral ABCD

∠ABC + ∠ADC = 180° (opposite angles in cyclic quadrilateral are supplementary) –2

From –1 and –2

∠ABF + ∠AEF = ∠ABC + ∠ADC

∠AEF = ∠ADC (∠ABF = ∠ABC)

Since these are corresponding angles

We can say that EF || DC

∴ EF||DC

Hence proved.