Prove that the quadrilateral formed by internal angle bisectors of a cyclic quadrilateral is also cyclic.
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Given: ABCD is a cyclic quadrilateral whose angle bisectors form the quadrilateral PQRS.
To Prove: PQRS is a cyclic
Proof: ABCD is a cyclic quadrilateral ∴∠A +∠C = 180° and ∠B+ ∠D = 180°
½ ∠A+½ ∠C = 90° and ½ ∠B+½ ∠D = 90°
x + z = 90° and y + w = 90°
In ΔARB and ΔCPD, x+y + ∠ARB = 180° and z+w+ ∠CPD = 180°
∠ARB = 180° – (x+y) and ∠CPD = 180° – (z+w)
∠ARB+∠CPD = 360° – (x+y+z+w) = 360° – (90+90)
= 360° – 180° ∠ARB+∠CPD = 180°
∠SRQ+∠QPS = 180°
The sum of a pair of opposite angles of a quadrilateral PQRS is 180°. Fig
Hence PQRS is cyclic
To Prove: PQRS is a cyclic
Proof: ABCD is a cyclic quadrilateral ∴∠A +∠C = 180° and ∠B+ ∠D = 180°
½ ∠A+½ ∠C = 90° and ½ ∠B+½ ∠D = 90°
x + z = 90° and y + w = 90°
In ΔARB and ΔCPD, x+y + ∠ARB = 180° and z+w+ ∠CPD = 180°
∠ARB = 180° – (x+y) and ∠CPD = 180° – (z+w)
∠ARB+∠CPD = 360° – (x+y+z+w) = 360° – (90+90)
= 360° – 180° ∠ARB+∠CPD = 180°
∠SRQ+∠QPS = 180°
The sum of a pair of opposite angles of a quadrilateral PQRS is 180°. Fig
Hence PQRS is cyclic
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