$A B C D$ is a quadrilateral inscribed in a circle. $C D$ is produced to any point $F$ and the bisector of $∠ A B C$ intersects the circle at $E$ . Prove that $D E$ bisects $∠ A D F$ .

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Solution

In quadrilateral $B C D E$
$∠ b = ∠ d$ (exterior angle of a cyclic quadrilateral is equal to the opposite interior angle)
In quadrilateral $A B C D$
$∠ B = ∠ A D F$ (exterior angle of a cyclic quadrilateral is equal to the opposite interior angle)
$⟹ ∠ a + ∠ b = ∠ c + ∠ d$
$⟹ ∠ a + ∠ b = ∠ c + ∠ b$
$∠ a = ∠ c$
and it is given that
$∠ a = ∠ b$
so, $∠ b = ∠ c$
Hence $∠ c = ∠ d$
$⟹ D E$ bisects $∠ A D F$

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