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 OB=OC​

Answer 1

Step-by-step explanation:

(i) In △ABC, we have

AB=AC

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

⇒ 1∠B= 1∠C

2 2

⇒∠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

⇒∠BAO=∠CAO ∣ Since corresponding parts of congruent triangles are equal

⇒ AO bisects ∠A.

Answer 2

Step-by-step explanation:

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Solution:-

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.

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