Question

In a triangle ABC, AB>AC and O is a point on AB such that AO=AC. Prove that (I) 2(angle ACO)=angle B+ angle C. (ii)angle BCO= ½(angle C- angle B).​

Answer 1

Explanation:

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.

Hint ↑

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