Question

Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any

quadrilateral is cyclic
​

Answer 1

Answer:

Hence Proved

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 + $\frac{1}{2}$∠A + $\frac{1}{2}$∠ B = 180°

∠AEB  = 180° –  $\frac{1}{2}$ (∠A + ∠ B) -------- (2)

From (1) and (2),

∠HEF = 180° –$\frac{1}{2}$(∠A + ∠ B) --------- (3)

Similarly, ∠HGF = 180° – $\frac{1}{2}$(∠C + ∠ D) -------- (4)

From 3 and 4,

∠HEF + ∠HGF = 360° –  $\frac{1}{2}$(∠A + ∠B + ∠C + ∠ D)

                         = 360° – $\frac{1}{2}$(360°)

                         = 360° – 180°

                         = 180°

So, EFGH is a cyclic quadrilateral [since the sum of the opposite

angles of the quadrilateral is 180°.]

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Answer 2

Refers to the attachment

Hope it helps u