The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.

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Solution

Given: $ABCD$ is a cyclic quadrilateral and $PQRS$ is a quadrilateral

In $\triangle APD$ :

$PAD + ADP + APD = 180^o$ … … … … … (i)

In $\triangle BQC$

$QBC + BCQ + BQC = 180^o$ … … … … … (ii)

Adding (i) and (ii)

$PAD + ADP + APD + QBC + BCQ + BQC = 360^o$

$\frac{1}{2} (BAD + ADC + DCB + CBA) + APD + BQC = 360^o$

$180 + APD + BQC = 360^o \Rightarrow APD + BQC = 180^o$

Therefore $PQRS$ is a cyclic quadrilateral as the opposite angles are supplementary.

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