Answer:
See Below.
Step-by-step explanation:
We are given the isosceles triangle ΔABC. By the definition of isosceles triangles, this means that ∠ABC = ∠ACB.
Segments BO and CO bisects ∠ABC and ∠ACB.
And we want to prove that ΔBOC is an isosceles triangle.
Since BO and CO are the angle bisectors of ∠ABC and ∠ACB, respectively, it means that ∠ABO = ∠CBO and ∠ACO = ∠BCO.
And since ∠ABC = ∠ACB, this implies that:
∠ABO = ∠CBO =∠ACO = ∠BCO.
This is shown in the figure as each angle having only one tick mark, meaning that they are congruent.
So, we know that:
![\angle ABC=\angle ACB](https://tex.z-dn.net/?f=%5Cangle%20ABC%3D%5Cangle%20ACB)
∠ABC is the sum of the angles ∠ABO and ∠CBO. Likewise, ∠ACB is the sum of the angles ∠ACO and ∠BCO. Hence:
![\angle ABO+\angle CBO =\angle ACO+\angle BCO](https://tex.z-dn.net/?f=%5Cangle%20ABO%2B%5Cangle%20CBO%20%3D%5Cangle%20ACO%2B%5Cangle%20BCO)
Since ∠ABO =∠ACO, by substitution:
![\angle ABO+\angle CBO =\angle ABO+\angle BCO](https://tex.z-dn.net/?f=%5Cangle%20ABO%2B%5Cangle%20CBO%20%3D%5Cangle%20ABO%2B%5Cangle%20BCO)
Subtracting ∠ABO from both sides produces:
![\angle CBO=\angle BCO](https://tex.z-dn.net/?f=%5Cangle%20CBO%3D%5Cangle%20BCO)
So, we've proven that the two angles are congruent, thereby proving that ΔBOC is indeed an isosceles triangle.