Question

In a triangle ABC, the internal bisectors of angle B and angle C meet at O. Prove that OA is also internal bisector of angle A.
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Answer 1

Answer:

ANSWER

Since the angles opposite to equal sides are equal,

∴AB=AC

⇒∠C=∠B

⇒

2

∠B

=

2

∠C

.

Since BO and CO are bisectors of ∠B and ∠C, we also have

∠ABO=

2

∠B

and ∠ACO=

2

∠C

.

∠ABO=

2

∠B

=

2

∠C

=∠ACO.

Consider △BCO:

∠OBC=∠OCB

⇒BO=CO ....... [Sides opposite to equal angles are equal]

Finally, consider triangles ABO and ACO.

BA=CA ...... (given);

BO=CO ...... (proved);

∠ABO=∠ACO (proved).

Hence, by S.A.S postulate

△ABO≅△ACO

⇒∠BAO=∠CAO⇒AO bisects ∠A.

Step-by-step explanation:

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