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 quadrilateralSum of angles of Quadrilateral = 360degress∠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 EFGHSince sum of one pair of opposite angles is 180 degrees