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If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S, then PQRS is a

(A) rectangle

(B) rhombus

(C) parallelogram

(D) quadrilateral whose opposite angles are supplementary

1 Answer

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Best answer

(D) quadrilateral whose opposite angles are supplementary

Explanation:

We know that,

Sum of all angles of a quadrilateral = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

Dividing LHS and RHS by 2,

⇒ ½ (∠A + ∠B + ∠C + ∠D) = ½ × 360° = 180°

Since, AP, PB, RC and RD are bisectors of ∠A, ∠B, ∠C and ∠D

⇒ ∠PAB + ∠ABB + ∠RCD + ∠RDC = 180° … (1)

We also know that,

Sum of all angles of a triangle = 180°

∠PAB + ∠APB + ∠ABP = 180°

⇒ ∠PAB + ∠ABP = 180° – ∠APB …(2)

Similarly,

∴ ∠RDC + ∠RCD + ∠CRD = 180°

⇒ ∠RDC + ∠RCD = 180° – ∠CRD …(3)

Substituting the value of equations (2) and (3) in equation (1),

180° – ∠APB + 180° – ∠CRD = 180°

⇒ 360° – ∠APB – ∠CRD = 180°

⇒ ∠APB + ∠CRD = 360° – 180°

⇒ ∠APB + ∠CRD = 180° …(4)

Now,

∠SPQ = ∠APB [vertically opposite angles]

∠SRQ = ∠DRC [vertically opposite angles]

Substituting in equation (4),

⇒ ∠SPQ + ∠SRQ = 180°

Hence, PQRS is a quadrilateral whose opposite angles are supplementary.

Therefore, option (D) is the correct answer.

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