Prove that the quadrilateral formed by the bisectors of the angles of a cyclic quadrilateral is also a cyclic quadrilateral.

Given that ABCD is a cyclic quadrilateral.

We have to prove that EFGH is also a cyclic quadrilateral.

Proof:

Let

⇒ 1/2∠A = x; 1/2∠B = w; 1/2∠C = z and 1/2∠D = y

We know that opposite angles of a cyclic quadrilateral are supplementary.

⇒∠A + ∠C = 180° and ∠B + ∠D = 180°

⇒ 1/2 (A + C) = 90° and 1/2 (B + D) = 180°

⇒∠x + ∠z = 90° and ∠y + ∠w = 90° … (1)

In ΔAFD and ΔBHC,

⇒∠x + ∠y + ∠AFD = 180°

⇒∠AFD = 180° - (∠x + ∠y) … (2)

And ∠z + ∠w + ∠BHC = 180°

⇒∠BHC = 180° - (∠z + ∠w) … (3)

Adding (2) and (3),

⇒∠AGD + ∠BHC = 360° - (∠x + ∠y + ∠z + ∠w)

From (1),

⇒∠AGD + ∠BHC = 360° - 180°

∴ ∠AGD + ∠BHC = 180°

⇒∠FGH + ∠HEF = 180° [Vertically opposite angles]

We know that opposite angles of a cyclic quadrilateral are supplementary.

∴ EFGH is also a cyclic quadrilateral.

Hence proved