Question

In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that
(i) OB = OC
(ii) AO bisects ∠A

Answer 1

Step-by-step explanation:

given AB=AC ................1

OB is the bisector of angle B

so angle ≤ABO = ≤OBC= 1/2 ≤B .................2

OC is the bisector of ≤C

so ≤ACO = ≤OCB = 1/2≤C................3

to prove OB=OC.

Proof:-

AB= AC

=angle ACB= angle ABC (angle opposite to equal sides are equal) to

1/2<ACB=1/2 < ABC

<OBC = <OCB. ( from EQ. 2 and 3)

hence OB = OC ( sides opposite to equal angles are equal)

hence proved

Answer 2

Given:

AB = AC and

the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

∠B = ∠C

½ ∠B = ½ ∠C

⇒ ∠OBC = ∠OCB (Angle bisectors)

∴ OB = OC (Side opposite to the equal angles are equal)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved Already)

So, ΔAOB ≅ ΔAOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects ∠A.

Hope it helpss !!