The bisectors of the equal angles B and C of an isosceles triangle ABC meet at O. Prove that AO bisects angle A.

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Solution

$Δ$ ABD is a isosceles triangle
AB = AC
CO bisects $∠$ C
BO bisects $∠$ B

(i) To be shown,
OB = OC

as, CO is the bisector of $∠$ C
$∠$ ACO = $∠$ OCB

and as, BO is the bisector of $∠$ B
$∠$ ABO = $∠$ OBC

$∴$ $∠$ OCB = $∠$ OBC

$∴$ OC = OB
[If two opposite angles are equal so opposite side are equal]

(ii) To be shown,
$Δ$ AOC = $Δ$ AOB

AC = AB (given)
AO = AO (common side)
OC = OB { proved in (i) }
$∠$ OCA = $∠$ ABO

$∴$ by {SAS} congruence condition
$Δ$ AOC $≅$ $Δ$ AOB

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