Question

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.

Answer 1

Now, ABFE being a cyclic quadrilateral, we have:
∠ABF + ∠AEF = 180° ...(i)
Also, ABCD being a cyclic quadrilateral, we have:
∠ABC + ∠ADC = 180°
or ∠ABF + ∠ADC = 180° ...(ii)
From (i) and (ii), we have:
∠ABF + ∠AEF = ∠ABF + ∠ADC
⇒ ∠AEF = ∠ADC

∴ EF || DC (∵ Corresponding angles are equal)