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Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic?
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Prove that the quadrilateral formed by the internal angle bisectors of...
Let ABCD be a quadrilateral in which the angle bisectors AH, BF, CF & DH of internal ∠A, ∠B, ∠C & ∠D respectively form a quadrilateral EFGH. EFGH is cyclic quadrilateral.
we have to prove that sum of one pair of opposite angles of a quadrilateral is 180degree .i.e ∠E+ ∠G = 180degree  OR ∠F +∠H= 180degree 
In ∆ AEB,
∠ABE + ∠ BAE + ∠AEB = 180degree 
∠AEB = 180degree - ∠ABE - ∠BAE
∠AEB = 180degree - (1/2 ∠B +1/2 ∠A)
∠AEB = 180degree  -1/2 (∠B + ∠A) 
[AH & BF are bisectors of ∠A & ∠B)
Lines AH and BF intersect
So ∠ FEH = ∠AEB (vertically opposite angle)
∠ FEH = 180degree  - 1/2 ( ∠B + ∠ A )
Similarly,∠ FGH = ∠GCD
∠FGH = 180degree -1/2 (∠C + ∠ D)
On adding eqs. (2) and (3)
∠FEH + ∠FGH = 180degree  -1/2 (∠A + ∠B) +180degree  - 1/2 (∠C +∠D)
∠FEH + ∠FGH = 180degree  + 180degree  - 1/2 (∠A + ∠B + ∠C + ∠D)
∠ FEH + ∠FGH = 360degree  - 1/2 (∠A + ∠B + ∠C + ∠D)
∠FEH + ∠FGH = 360degree  -1/2 x 360degree 
[∠A + ∠ B + ∠c + ∠ D = 360degree , Sum of angles of Quadrilateral is 360degree ]
∠ FEH +∠ FGH = 360degree  - 180degree 
∠ FEH + ∠ FGH = 180degree 
Hence,EFGH is a cyclic quadrilateral in which the sum of one pair of opposite angles is 180degree i.e.∠ FEH + ∠ FGH = 180degree.

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Prove that the quadrilateral formed by the internal angle bisectors of any quadrilateral is cyclic?
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