Question

In an isosceles triangle ABC, AB=AC. If BO and CO, the bisector of angle B &C meet at O And BC is produced to D, prove that angle BOC =angle ACD

Answer 1

Hlo mate :-

Solution :-

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☆ ΔABC, AB = AC ⇒ ∠B = ∠C

[Angles opposite to equal sides are equal]

● Also OA and OB are bisectors of angles B and C.

 1/2angle a = 1/2 angle b

 ⇒ ∠OBC = ∠OCB ∴ OB = OC

☆[Sides opposite to equal angles are equal] 

●Now consider, Δ’s AOB and AOC 

:- OA = OA (Common side)
:-  AB = AC (Given) 
:- OB = OC (Proved) 

:- ΔAOB ≅ ΔAOC [By SSS congruence criterion]

 ⇒ ∠OAB = ∠OAC That is OA is bisector ∠A.

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☆ ☆ ☆ Hop It's helpful ☆ ☆ ☆

Answer 2

Here is your solution

Given :-

In Δ ABC

AB = AC 
∠B = ∠C

(Angles opposite to equal sides are equal)

Also OA and OB are bisectors of angles B and C.

 1/2 ∠ a = 1/2 ∠ b

 =>∠OBC = ∠OCB

OB = OC

(Sides opposite to equal angles are equal)

In Δ AOB and AOC 

OA = OA (Common side)
AB = AC (equal side) 
OB = OC (equal side ) 

ΔAOB ≅ ΔAOC [By SSS congruence criterion]

Hence

∠OAB = ∠OAC 

so,

OA is bisector ∠A.(proved)

hope it helps you