Question

Prove that bisector of internal angle of the quadriateral forms a cyclic quadriateral​

Answer 1

answer

Let ABCD be a quadrilateral in which the angle bisectors AH, BF, CF & DH of internal ∠A, ∠B, ∠C & ∠D respectively form a quadrilateral EFGH. ... Hence,EFGH is a cyclic quadrilateral in which the sum of one pair of opposite angles is 180° i.e.∠ FEH + ∠ FGH = 180°.

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Answer 2

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.

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