Bisector AD of $∠ B A C$ of $Δ A B C$ passes through the centre O of the circumcircle of $Δ A B C$ . Prove that AB = AC. [2 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark

Bisector AD of $∠ B A C$ of $Δ A B C$ passes through the centre O of the circumcircle of $Δ A B C$ .

To prove that AB = AC

Construction: Draw $O P ⊥ A B$ and $O Q ⊥ A C$

Proof :

In $Δ A P O$ and $Δ A Q O$

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

$∠ O A P = ∠ O A Q$ Given

$O A = O A$ Common

$∴ Δ A P O ≅ Δ A Q O$ by AAS

$∴ O P = O Q$ by C.P.C.T

$∴ A B = A C$ Two chords equidistant from the center are equal.

Hence proved.

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In $Δ$ ABC, AB = AC and the bisectors of angles B and C intersect at point O. Prove that BO = CO and the ray AO is the bisector of angle BAC.
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Bisector AD of ∠BAC of ΔABC passes through the centre O of the circumcircle of ΔABC. Prove that AB = AC. [2 MARKS]

Bisector AD of $∠ B A C$ of $Δ A B C$ passes through the centre O of the circumcircle of $Δ A B C$ . Prove that AB = AC. [2 MARKS]