Q. 13

# Prove that the bisectors of the angles, formed by producing opposite sides (which are not parallel) of a cyclic quadrilateral, intersect each other at right-angle.

Answer :

Let P be a point on a circumcircle of ΔABC. Perpendiculars PL, PM and PN are drawn on the lines through the line segments BC, CA and AB respectively.

ABCD is a cyclic quadrilateral (Fig.) AD and BC when produced meet at P. DC and AB, when produced, meet at O. Angle bisectors of ∠P and ∠O meet at R.

To Prove: ∠PRO = 90°

Construction: Produce PR to meet AB in S.

Proof: In ΔPDL and ΔPSB,

∠DPL = ∠BPS (given)

∠PDL = ∠PBS (exterior angle of a cyclic quadrilateral)

.

But ∠PLD = ∠OLR (Vertically Opposite angles)

⇒ ∠OLR = ∠PSB = ∠RSO

Now in ΔOLR and ΔOSR

∠ROL = ∠ROS (given)

∠RLO = ∠RSO (proved)

∴ ∠ORL = ∠ORS

But ∠ORL + ∠ORS = 180°

∴ ∠ORL = 90°

⇒ ∠PRO = 90°

ABCD is a cyclic quadrilateral (Fig.) AD and BC when produced meet at P. DC and AB, when produced, meet at O. Angle bisectors of ∠P and ∠O meet at R.

To Prove: ∠PRO = 90°

Construction: Produce PR to meet AB in S.

Proof: In ΔPDL and ΔPSB,

∠DPL = ∠BPS (given)

∠PDL = ∠PBS (exterior angle of a cyclic quadrilateral)

.

^{.}. ∠PLD = ∠PSBBut ∠PLD = ∠OLR (Vertically Opposite angles)

⇒ ∠OLR = ∠PSB = ∠RSO

Now in ΔOLR and ΔOSR

∠ROL = ∠ROS (given)

∠RLO = ∠RSO (proved)

∴ ∠ORL = ∠ORS

But ∠ORL + ∠ORS = 180°

∴ ∠ORL = 90°

⇒ ∠PRO = 90°

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