Question

In an isosceles triangle ABC, with AB=AC, the bisectors of angle B and angle c intersect
each other at O. Join A to 0. Show that OB=OC.​

Answer 1

Answer:

Step-by-step explanation:

In ΔABC, AB = AC

⇒ ∠B = ∠C [Angles opposite to equal sides are equal]

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

⇒ ∠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.

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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