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