Math, asked by deepanjalipande67, 1 year ago

prove that the quadrilateral formed by bisectors of angles of any quadrilateral is a cyclic quadrilateral.


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Answers

Answered by onlinewithaalia
8

ABCD is a cyclic quadrilateral

∠A+∠C=180° and ∠B+∠D = 180°

(∠A+∠C)/2=90°and(∠B+∠D)/2= 90°

w,x,y,z are angles of the inner quadrilateral

x + z = 90°

y + w = 90°

In ΔAGD and ΔBEC,

x + y + ∠AGD = 180°

z + w + ∠BEC = 180°

∠AGD = 180° – (x+y)

∠BEC = 180° – (z+w)

∠AGD+∠BEC=360°-(x+y+z+w)

= 360° – (90+90) = 360° – 180° = 180°

∠AGD+∠BEC = 180°

∠FGH+∠HEF = 180°

The sum of a pair of opposite angles of a quadrilateral EFGH is 180°.

Hence EFGH is cyclic

Hence Proved

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Answered by mathdude500
0

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

Let assume that ABCD be a 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. \\  \\

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