Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.

(N/A) In the figure,$ABCD$ is a quadrilateral in which the angle bisectors of internal angles $A, B, C$ and $D$ intersect to form a quadrilateral $EFGH$.
Now,in $\triangle AEB$,$\angle FEH = \angle AEB = 180^{\circ} - (\angle EAB + \angle EBA)$
$= 180^{\circ} - \frac{1}{2}(\angle A + \angle B)$
Similarly,in $\triangle CGD$,$\angle FGH = \angle CGD = 180^{\circ} - (\angle GCD + \angle GDC)$
$= 180^{\circ} - \frac{1}{2}(\angle C + \angle D)$
Adding these two equations:
$\angle FEH + \angle FGH = 180^{\circ} - \frac{1}{2}(\angle A + \angle B) + 180^{\circ} - \frac{1}{2}(\angle C + \angle D)$
$= 360^{\circ} - \frac{1}{2}(\angle A + \angle B + \angle C + \angle D)$
Since the sum of angles in a quadrilateral is $360^{\circ}$,we have:
$= 360^{\circ} - \frac{1}{2}(360^{\circ}) = 360^{\circ} - 180^{\circ} = 180^{\circ}$
Since the sum of opposite angles of quadrilateral $EFGH$ is $180^{\circ}$,it is a cyclic quadrilateral.