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.
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)
... ∠PLD = ∠PSB
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 = ∠PSB
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°
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