AB and CD are the chords of a circle whose center is O. They intersect each other at P. If PO is the bisector of $∠ A P D$ , prove that AB = CD. [2 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark

Given :AB and CD are the chords of a circle whose center is O.
They intersect each other at P.
PO is the bisector of $∠ A P D$

To prove: AB = CD

Construction: Draw $O R ⊥ A B$ and $O Q ⊥ C D$ .

Proof: In $Δ O P R$ and $Δ O P Q$

$∠ O P R = ∠ O P Q$ (Given)

$O P = O P$ (Common)

$∠ O R P = ∠ O Q P$ (Each $= 90^{\circ}$ by construction)

$∴ Δ O P R = Δ O P Q$ (AAS congruency)

$∴ O R = O Q$ (C.P.C.T)

$∴ A B = C D$ ( $∵$ chord of a circle which are equidistant from the center are equal).

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AB and CD are the chords of a circle whose center is O. They intersect each other at P. If PO is the bisector of ∠APD, prove that AB = CD. [2 MARKS]

AB and CD are the chords of a circle whose center is O. They intersect each other at P. If PO is the bisector of $∠ A P D$ , prove that AB = CD. [2 MARKS]