Question

Prove that the quadrilateral formed by the bisector of another quadrilateral is cyclic?

Answer 1

$\mathfrak{\huge{\orange{\underline{\underline{Answer}}}}}$

Given: A cyclic quadrilateral ABCD in which the angle bisectors AR, BR, CP O and DP of internal angles A, B, C and D respectively form a quadrilateral PQRS.

To prove: PQRS is a cyclic quadrilateral

Proof: In △ARB, we have

1/2∠A + 1/2∠B + ∠R = 180° ....(i) (Since, AR, BR are bisectors of ∠A and ∠B)

In △DPC, we have

1/2∠D + 1/2∠C + ∠P = 180° ....(ii) (Since, DP,CP are bisectors of ∠D and ∠C respectively)

Adding (i) and (ii),we get

1/2∠A + 1/2∠B + ∠R +1/2∠D + 1/2∠C + ∠P = 180° + 180°

∠P + ∠R = 360° - 1/2(∠A + ∠B + ∠C + ∠D)

∠P + ∠R = 360° - 1/2 x 360° = 360° - 180°

⇒ ∠P + ∠R = 180°

As the sum of a pair of opposite angles of quadrilateral PQRS is 180°. Therefore, quadrilateral PQRS is cyclic.

I have explained this question with sum also

So,I hope it helps you mate✌️

Answer 2

Answer:

Given: A cyclic quadrilateral ABCD in which the angle bisectors AR, BR, CP O and DP of internal angles A, B, C and D respectively form a quadrilateral PQRS. To prove: PQRS is a cyclic quadrilateral.