Question

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

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Answer 1

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Prove that the quadrilateral formed by the internal angle bisector of any quadrilateral formed is cyclic.

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Let ABCD is a quadrilateral.

AH,BF,CF,DH

are bisector of ∠A,∠B,∠C,∠D.

EFGH is also cyclic quadrilateral.

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We have to prove that,

Sum pair of opposite angle is 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 and BF are bisectors of ∠A and ∠B ]

∠FEH=∠AEB

∠FEH=180°-1/2 ( ∠B+ ∠A)---(2)

∠FEH=180° −1/2(∠B+∠A)−−−(2)

Similarly,

∠FGH=180°-1/2(∠C+∠D)......(3)

By adding eq. (2)&(3)

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

∠FEH+∠FGH=180°- 1/2(∠A+∠B+∠C+∠D)

Since,ABCD is quadrilateral.

Sum of angles= 360°=(∠A+∠B+∠C+∠D)

∠FEH+∠FGH=360°-1/2(360°)=180°

Thus EFGH is cyclic quadrilateral.

Since sum of one pair of opposite angle is=180°.

______________________Proved.