Question

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

Answer 1

Let us consider $ABCD$ is a quadrilateral $AH,BF,CF,DH$ are bisector of $\angle A,\angle B,\angle C,\angle D$ we have to prove $EFGH$ is cyclic quadrilateral to prove $EFGH$ is cyclic we have to prove sum of one pair of opposite angle is ${180}^o$

In $△le$ $AEB\to \angle ABE+\angle BAE+\angle AEB={180}^o$

$\angle AEB={180}^o-\angle ABE-\angle BAE$

$\angle AEB={180}^o(1/2\angle B+1/2\angle A)={180}^o-1/2(\angle B+\angle A)---\left(1\right)$

Now lines $AH$ and $BF$ intersect

So $\angle FEH=\angle AEB$

$\angle FEH={180}^o-1/2(\angle B+\angle A)---\left(2\right)$

Similarly $\angle FGH={180}^o-\frac{1}{2}(\angle C+\angle D)---\left(3\right)$

Add $\left(2\right)$ & $\left(3\right)$ $\angle FEH+\angle FGH={180}^o-1/2(\angle B+\angle C)+{180}^o-1/2(\angle A+\angle D)$

$\angle FEH+\angle FGH={180}^o+{180}^o-1/2(\angle A+\angle B+\angle C+\angle D)$

Since $ABCD$ is quadrilateral

Sum of angles $=360=\angle A+\angle B+\angle C+\angle D$

$\angle FEH+\angle FGH=360-\frac{1}{2}\left(360\right)=360-180={180}^o$

$\angle FEH+\angle FGH={180}^o$

Thus $EFGH$ is cyclic quadrilateral.

Since sum of one pair of opposite angle is ${180}^o$.