Prove that the quadrilateral formed by angle bisectors of a cyclic
quadrilateral is also cyclic.
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Here, ABCD is a cyclic quadrilateral.
AH,BF,CF and DH are the angle bisectors of ∠A,∠B,∠C and ∠D.
⇒ ∠FEH=∠AEB --- ( 1 ) [ Vertically opposite angles ]
⇒ ∠FGH=∠DGC ---- ( 2 ) [ Vertically opposite angles ]
Adding ( 1 ) and ( 2 ),
⇒ ∠FEH+∠FGH=∠AEB+∠DGC --- ( 3 )
Now, by angle sum property of a triangle,
⇒ ∠AEB=180
o
−(
2
1
∠A+
2
1
∠B) ---- ( 4 )
⇒ ∠DGC=180
o
−(
2
1
∠C+
2
1
∠D) ---- ( 5 )
Substituting ( 4 ) and ( 5 ) in equation ( 3 )
⇒ ∠FEH+∠FGH=180
o
−(
2
1
∠A+
2
1
∠B)+180
o
−(
2
1
∠C+
2
1
∠D)
⇒ ∠FEH+∠FGH=360
o
−
2
1
(∠A+∠B+∠C+∠D)
⇒ ∠FEH+∠FGH=360
o
−
2
1
×360
o
⇒ ∠FEH+∠FGH=180
o
Now, the sum of opposite angles of quadrilateral EFGH is 180
o
.
∴ EFGH is a cyclic quadrilateral.
Hence, the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
AH,BF,CF and DH are the angle bisectors of ∠A,∠B,∠C and ∠D.
⇒ ∠FEH=∠AEB --- ( 1 ) [ Vertically opposite angles ]
⇒ ∠FGH=∠DGC ---- ( 2 ) [ Vertically opposite angles ]
Adding ( 1 ) and ( 2 ),
⇒ ∠FEH+∠FGH=∠AEB+∠DGC --- ( 3 )
Now, by angle sum property of a triangle,
⇒ ∠AEB=180
o
−(
2
1
∠A+
2
1
∠B) ---- ( 4 )
⇒ ∠DGC=180
o
−(
2
1
∠C+
2
1
∠D) ---- ( 5 )
Substituting ( 4 ) and ( 5 ) in equation ( 3 )
⇒ ∠FEH+∠FGH=180
o
−(
2
1
∠A+
2
1
∠B)+180
o
−(
2
1
∠C+
2
1
∠D)
⇒ ∠FEH+∠FGH=360
o
−
2
1
(∠A+∠B+∠C+∠D)
⇒ ∠FEH+∠FGH=360
o
−
2
1
×360
o
⇒ ∠FEH+∠FGH=180
o
Now, the sum of opposite angles of quadrilateral EFGH is 180
o
.
∴ EFGH is a cyclic quadrilateral.
Hence, the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
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