Question

16. a) In the given figure, OB bisects ABC and OC bisects ACB, prove that BOC - BAC = ABO+ACO.​

Answer 1

We have shown that ∠BOC - ∠BAC = ∠ABO + ∠ACO, as required.

Let's start by considering the angles in the figure described.

Given: OB bisects ∠ABC, so ∠OBA = ∠OBC (angles are equal).

Also, OC bisects ∠ACB, so ∠OCA = ∠OCB (angles are equal).

Now, let's calculate the angles:

We want to prove that ∠BOC - ∠BAC = ∠ABO + ∠ACO.

From the angle bisector property, we know that ∠OBA + ∠OBC = ∠ABC, and ∠OCA + ∠OCB = ∠ACB.

So, we can rewrite ∠ABC and ∠ACB in terms of the angles formed by the bisectors:

∠ABC = ∠OBA + ∠OBC

∠ACB = ∠OCA + ∠OCB

Now, we can express ∠BOC and ∠BAC in terms of these angles:

∠BOC = ∠OBC + ∠OCB (Angles around a point sum up to 180°)

∠BAC = ∠OBA + ∠OCA (Angles around a point sum up to 180°)

Substitute the expressions we found earlier for ∠ABC and ∠ACB into these equations:

∠BOC = ∠ABC (substitute from above)

∠BAC = ∠ACB (substitute from above)

Now, let's rewrite our original equation:

∠BOC - ∠BAC = (∠ABC) - (∠ACB)

= (∠OBA + ∠OBC) - (∠OCA + ∠OCB)

= ∠OBA + ∠OBC - ∠OCA - ∠OCB

= ∠OBA + ∠OBC + ∠OCA + ∠OCB - 2∠OCB

= (∠OBA + ∠OCA) + (∠OBC + ∠OCB) - 2∠OCB

= ∠ABO + ∠ACO

Therefore, we have shown that ∠BOC - ∠BAC = ∠ABO + ∠ACO, as required.

Answer 2

Points ABC and points OBC form two separate triangles. Triangle ABC is made up of angles BAC, ABC, ACB, while triangle OBC is made up of angles BOC, OBC, OCB. Since interior angles of triangles add up to 180 degrees, m<BAC + m<ABC + m<ACB = 180 and m<BOC + m<OBC + m<OCB = 180. 180 is equal to 180, so m<BAC + m<ABC + m<ACB = m<BOC + m<OBC + m<OCB.

Since OB bisects <ABC and OC bisects <ACB, m<ABO = m<OBC and m<ACO = m<OCB. We can also say that m<ABC = 2m<ABO and m<ACB = 2m<ACO. Substitute and simplify the equation.

m<BAC + m<ABC + m<ACB = m<BOC + m<OBC + m<OCB

m<BAC + 2m<ABO + 2m<ACO = m<BOC + m<ABO + m<ACO

m<ABO + m<ACO = m<BOC - m<BAC