Question

prove that quadrilateral formed by the internal bisector of any quadrilateral is cyclic ​

Answer 1

Answer:

Step-by-step explanation:

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°.]

Answer 2

Answer:

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°.]