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Question

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


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,
ΔAOC = ΔAOB

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

by {SAS} congruence condition
ΔAOC ΔAOB

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