Question

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 quadrilateral whose opposite angles are __________.

Answer 1

Given:
ABCD is a quadrilateral
bisectors of ∠A and ∠B intersect each other at P
bisectors of ∠B and ∠C intersect each other at Q
bisectors of ∠C and ∠D intersect each other at R
bisectors of ∠D and ∠A intersect each other at S

In ∆DAS,
∠ASD + ∠SDA + ∠DAS = 180° (angle sum property)
⇒ ∠ASD + $\frac{1}{2}$∠D + $\frac{1}{2}$∠A = 180°
⇒ ∠ASD = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A

Also, ∠PSR = ∠ASD = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A
⇒ ∠PSR = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A ...(1)

Similarly,
In ∆BQC,
∠BQC + ∠QCB + ∠CBQ = 180° (angle sum property)
⇒ ∠BQC + $\frac{1}{2}$∠C + $\frac{1}{2}$∠B = 180°
⇒ ∠BQC = 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B

Also, ∠PQR = ∠BQC = 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B
⇒ ∠PQR = 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B ...(2)

Adding (1) and (2), we get
∠PSR + ∠PQR = 180° − $\frac{1}{2}$∠D − $\frac{1}{2}$∠A + 180° − $\frac{1}{2}$∠C − $\frac{1}{2}$∠B
= 360° − $\frac{1}{2}$(∠D + ∠A + ∠C + ∠B)
= 360° − $\frac{1}{2}$(360°)
= 360° − 180°​
= 180°

Since, sum of angles of a quadrilateral is 360°
Therefore, ∠PSR + ∠PQR + ∠QRS + ∠QPS = 360°
⇒ ∠QRS + ∠QPS = 180°

Hence, the sum of the opposite angles of the quadrilateral PQRS is 180°.

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