Question

in an isosceles triangle ABC with ab is equals to see the bisector of angle b and c intersect each other at point a to show that​

Answer 1

Answer:

∠C=∠B ∣ Since angles opposite to equal sides are equal

⇒21∠B=21∠C

⇒∠OBC=∠OCB

⇒∠ABO=∠ACO    …(1)

⇒OB=OC ∣ Since sides opp. to equal ∠s are equal     …(2)

(ii) Now, in △ABO and △ACO, we have 

AB=AC ∣ Given

∠ABO=∠ACO ∣ From (1)

OB=OC ∣ From (2)

∴ By SAS criterion of congruence, we have 

△ABO≅△ACO

Step-by-step explanation:

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Answer 2

Answer:

Given:

AB = AC.

BO is bisector of angle B.

CO is bisector of angle C.

To prove: BO = CO.

Proof:

Therefore, angle 1 = angle 2 and

angle 3 = angle 4.

In ∆ABC,

AB = AC (given)

=> angle A = angle C.

(angles opposite to equal sides are equal)

$=> \frac{angle A}{2} = \frac{angle C}{2}$

(Dividing both sides by 2)

=> angle 1 = angle 3 = angle 2 = angle 4

(angle 1 = angle 2)

As in ∆BOC, angle 2 = angle 4

BO = CO

(sides opposite to equal angles are equal)

Hence Proved.

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