In △ ABC, AB = AC and the bisectors ∠ B and ∠ C meet at a point

1.	In △ ABC, AB = AC and the bisectors ∠ B and ∠ C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠ A.
Answer» It is given that AB = AC and the bisectors ∠ B and ∠ C meet at a point O Consider △ BOC So we get ∠ BOC = ½ ∠ B and ∠ OCB = ½ ∠ C It is given that AB = AC so we get ∠ B = ∠ C So we get ∠ OBC = ∠ OCB We know that if the base angles are equal even the sides are equal So we get OB = OC ……. (1) ∠ B and ∠ C has the bisectors OB and OC so we get ∠ ABO = ½ ∠ B and ∠ ACO = ½ ∠ C So we get ∠ ABO = ∠ ACO …….. (2) Considering △ ABO and △ ACO and equation (1) and (2) It is given that AB = AC By SAS congruence criterion △ ABO ≅ △ ACO ∠ BAO = ∠ CAO (c. p. c. t) Therefore, it is proved that BO = CO and the ray AO is the bisector of ∠ A.

In △ ABC, AB = AC and the bisectors ∠ B and ∠ C meet at a point O. Prove that BO = CO and the ray AO is the bisector of ∠ A.

Answer»

It is given that AB = AC and the bisectors ∠ B and ∠ C meet at a point O

Consider △ BOC

So we get

∠ BOC = ½ ∠ B and ∠ OCB = ½ ∠ C

It is given that AB = AC so we get ∠ B = ∠ C

So we get

∠ OBC = ∠ OCB

We know that if the base angles are equal even the sides are equal

So we get OB = OC ……. (1)

∠ B and ∠ C has the bisectors OB and OC so we get

∠ ABO = ½ ∠ B and ∠ ACO = ½ ∠ C

So we get

∠ ABO = ∠ ACO …….. (2)

Considering △ ABO and △ ACO and equation (1) and (2)

It is given that AB = AC

By SAS congruence criterion

△ ABO ≅ △ ACO

∠ BAO = ∠ CAO (c. p. c. t)

Therefore, it is proved that BO = CO and the ray AO is the bisector of ∠ A.