Question

prove that the quadrilateral formed by angle bisector of a cyclic quadrilateral is also cyclic

Answer 1

as we know that in a cyclic quad the some of two opposite angles is 180 degree
then this would apply to all the half angles
so in triangle ABR AND DPC
angle R + P= 180
PROOF

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that ABCD be a cyclic quadrilateral and let further assume that the quadrilateral formed by angle bisectors of angle A, angle B, angle C and angle D be PQRS.

Let assume that

$\sf \: \angle A = 2x, \: \angle B = 2y, \: \: \angle C = z, \: \: \angle D = w$

Now, We know, sum of interior angles of a quadrilateral is 360°.

So, using this property, we have

$\sf \:\angle A + \angle B + \angle C + \angle D = {360}^{ \circ}$

$\sf \:\dfrac{1}{2} \angle A +\dfrac{1}{2} \angle B +\dfrac{1}{2} \angle C + \dfrac{1}{2}\angle D = {180}^{ \circ}$

$\implies\sf \:\boxed{\sf \: x + y + z + w= {180}^{ \circ} \: } - - - (1)$

Now, In triangle ARB

We know, sum of interior angles of a triangle is 180°.

So, using this property, we have

$\sf \: \angle A + \angle R + \angle B = {180}^{ \circ}$

$\implies\sf \: \boxed{\sf \: x + y + \angle B = {180}^{ \circ} \: } - - - (2)$

Now, In triangle CPD

We have

$\sf \: \angle C + \angle P + \angle D = {180}^{ \circ}$

$\sf \: z + \angle P +w = {180}^{ \circ}$

$\implies\sf \: \boxed{\sf \: z + w + \angle P = {180}^{ \circ} \: } - - - (3)$

On adding equation (2) and (3), we get

$\sf \: x + y + z + w + \angle P + \angle R = {360}^{ \circ}$

$\sf \: {180}^{ \circ} + \angle P + \angle R = {360}^{ \circ} \: \: \: \left[ \because \: of \: equation \: (1) \: \right]$

$\sf \:\angle P + \angle R = {360}^{ \circ} - {180}^{ \circ}$

$\implies\sf \: \angle P + \angle R = {180}^{ \circ}$

$\implies\sf \: PQRS \: is \: a \: cyclic \: quadrilateral.$