ABC is an isosceles triangle with the base AC, BO is the height lowered to the base. Prove that ΔABO and ΔСBО.

Before proving the equality of triangles ABO and CBO, you should pay attention to the first sign of equality of triangles.
According to the first sign of equality of triangles, if two sides and the angle between them of one triangles are respectively equal to two sides and the angle between them of another triangle, then these triangles are equal.
Consider these triangles.
1.) By hypothesis, an isosceles triangle ABC is given.
Hence, sides AB and BC of triangle ABC are equal.
AB = BC.
2.) Also, by condition, the BO side is common.
3.) Since, by condition, the side BO is the height in an isosceles triangle ABC, we recall the rule.
According to this rule, the height drawn in an isosceles triangle to the base is also a bisector.
Since it is known that the bisector divides the angle in half, we will write.
Angle ABO is equal to angle CBO at vertex B.
4.) Thus, in the considered triangles ABO and CBO, the sides AB and BC are equal, the side BO is common, and the angles between these sides are also equal, which means that the triangles ABO and CBO.