Question

prove that quad. formed by angle bisector of a cyclic quad. is also cyclic.

Answer 1

Hii dear here is your answer hope it helps you.......



Given a quadrilateral ABCD with internal angle bisectors AF, BH, CH
and DF of angles A, B, C and D respectively and the points E, F, G
and H form a quadrilateral EFGH.

To prove that EFGH is a cyclic quadrilateral.

∠HEF = ∠AEB [Vertically opposite angles] -------- (1)
Consider triangle AEB,
∠AEB + ½ ∠A + ½ ∠ B = 180°
∠AEB  = 180° – ½ (∠A + ∠ B) -------- (2)

From (1) and (2),
∠HEF = 180° – ½ (∠A + ∠ B) --------- (3)

Similarly, ∠HGF = 180° – ½ (∠C + ∠ D) -------- (4)

From 3 and 4,
∠HEF + ∠HGF = 360° – ½ (∠A + ∠B + ∠C + ∠ D)
                          = 360° – ½ (360°)
                          = 360° – 180°
                          = 180°
So, EFGH is a cyclic quadrilateral since the sum of the opposite
angles of the quadrilateral is 180°.]

hope it helps you....☺☺☺

Answer 2

Given a quadrilateral ABCD with internal angle bisectors AF, BH, CH and DF of angles A, B, C and D respectively and the points E, F, G and H form a quadrilateral EFGH. To prove that EFGH is a cyclic quadrilateral. ∠HEF = ∠AEB [Vertically opposite angles] -------- (1) Consider triangle AEB, ∠AEB + ½ ∠A + ½ ∠ B = 180° ∠AEB = 180° – ½ (∠A + ∠ B) -------- (2) From (1) and (2), ∠HEF = 180° – ½ (∠A + ∠ B) --------- (3) Similarly, ∠HGF = 180° – ½ (∠C + ∠ D) -------- (4) From 3 and 4, ∠HEF + ∠HGF = 360° – ½ (∠A + ∠B + ∠C + ∠ D) = 360° – ½ (360°) = 360° – 180° = 180° So, EFGH is a cyclic quadrilateral since the sum of the opposite angles of the quadrilateral is 180°.]