Question

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

Answer 1

Given: ABCD is a quadrilateral AH, BF, CF, DH are bisectors of ∠A, ∠B ∠C ∠D respectively

To prove: EFGH is cyclic quadrilateral

Proof: To prove EFGH is a cyclic quadrilateral, we prove that sum of one pair of opposite angles is 180 degrees.

in triangle AEB

∠ABE + ∠ BAE + ∠AEB = 180 degrees 

∠AEB = 180 degress - ∠ABE - ∠BAE

∠AEB = 180 degress (1/2 ∠B +1/2 ∠A)

∠AEB = 180 degrees -1/2 (∠B + ∠A) ....(1)

Now Lines AH and BF intersect

So ∠ FEH = ∠AEB

∠ FEH = 180 degrees - 1/2 ( ∠B + ∠ A ) ....(2)

Similarly we can prove that 
∠FGH = 180 degrees = 1/2 (∠C + ∠ D).... (3)

addding (2) and (3)

∠FEH + ∠ FGH = 180 degrees -1/2 (∠A + ∠D) +180 degress - 1/2 (∠C +∠ B)

∠FEH + ∠FGH = 180 degrees + 180 Degrees - 1/2 (∠A + ∠D + ∠C + ∠B)

∠ FEH + ∠FGH = 360 degrees - 1/2 (∠A + ∠B + ∠C + ∠D)

∠ FEH + ∠ FGH = 360 degrees -1/2 (∠A + ∠B +∠C + ∠D)

Since ABCD is a quadrilateral
Sum of angles of Quadrilateral = 360 degress
∠A + ∠ B + ∠c + ∠ D = 360 Degrees

∠FEH + ∠FGH = 360 degrees -1/2 x 360 degrees

∠ FEH + FGH = 360 degrees - 180 degrees

∠ FEH + ∠ FGH = 180 degrees

Thus in EFGH
Since sum of one pair of opposite angles is 180 degrees

EFGH is a cyclic quadrilateral

Answer 2

[FIGURE IS IN THE ATTACHMENT]
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 180°.i.e ∠E+ ∠G = 180° OR ∠F +∠H= 180°
In ∆ AEB,
∠ABE + ∠ BAE + ∠AEB = 180°
∠AEB = 180°- ∠ABE - ∠BAE
∠AEB = 180°- (1/2 ∠B +1/2 ∠A)
∠AEB = 180° -1/2 (∠B + ∠A) .............(1)
[AH & BF are bisectors of ∠A & ∠B)
Lines AH and BF intersect
So ∠ FEH = ∠AEB (vertically opposite angle)
∠ FEH = 180° - 1/2 ( ∠B + ∠ A ) ……....(2)
Similarly,∠ FGH = ∠GCD
∠FGH = 180°-1/2 (∠C + ∠ D)............... (3)
On adding eqs. (2) and (3)
∠FEH + ∠FGH = 180° -1/2 (∠A + ∠B) +180° - 1/2 (∠C +∠D)
∠FEH + ∠FGH = 180° + 180° - 1/2 (∠A + ∠B + ∠C + ∠D)
∠ FEH + ∠FGH = 360° - 1/2 (∠A + ∠B + ∠C + ∠D)
∠FEH + ∠FGH = 360° -1/2 x 360°
[∠A + ∠ B + ∠c + ∠ D = 360°, Sum of angles of Quadrilateral is 360°]
∠ FEH +∠ FGH = 360° - 180°
∠ FEH + ∠ FGH = 180°
Hence,EFGH is a cyclic quadrilateral in which the sum of one pair of opposite angles is 180° i.e.∠ FEH + ∠ FGH = 180°.

HOPE THIS WILL HELP YOU.....