Question

ABC is an Isosceles Triangle with AB = AC, the Bisector of ∠B and ∠C intersect each other at O. Join A to O.

Show that -
(i) OB = OC
(ii) AC bisects ∠A

Answer 1

EXPLAINATION

Given :- In triangle ABC

AB = AC

BO is the bisector of angle B.

Therefore, angle 1 = angle 2

== CO is the bisector of angle C.

Therefore , angle 3 = angle 4.

To proof :- i) OB = OC

ii) AO bisects angle A.

Proof :- In triangle ABC

i) AB = AC

ii) angle B = angle C.or angle 1 + angle 2 = angle 3 + angle

= (angle 1 = angle 2 and angle 3 = angle 4).

Therefore , 2 angle 2 = 2 Angle 4

=> angle 2 = angle 4

Therefore, OB = OC (sides opposite to equal angles are equal).

ii) In triangle ABO and triangle ACO.

AO = OA (common)

AB = AC (given)

OB = OC (from 3rd)

Therefore , ABO congruent ACO (SSS)

Angle 5 = angle 6 (cpct).

AO bisects angle A.

HENCE PROVED

Answer 2

EXPLANATION.

ΔABC is an isosceles triangle.

⇒ AB = AC.

Bisector ∠B and ∠C intersect each other at O.

Join A to O.

As we know that,

in isosceles triangle two sides are equal.

O is the mid-point of the triangle.

∠ABC = ∠ACB [Angle opposite to equal sides of a triangle is equal].

OB is a bisector of ∠B.

⇒ ∠ABO = ∠OBC = 1/2∠B ⇒ (1).

OC is a bisector of ∠C.

⇒ ∠ACO = ∠OCB = 1/2∠C ⇒ (2).

⇒ ∠OBC = ∠OCB.

Therefore,

⇒ OB = OC.

HENCE PROVED.

AC bisects ∠A.

In ΔABC = ΔACB.

⇒ AB = AC [ Given].

⇒ AC = AC [ Common].

⇒ ΔABC ≅ ΔACB [ SSS criteria ].

⇒ ∠ABC = ∠ACB [ C.P.C.T].

HENCE PROVED.