In the isosceles triangle ABC, the height BH is drawn. Prove that triangle ABH and BCH are equal.

You are given an isosceles triangle ABC.
The height BH is drawn in this triangle.
In order to prove that triangles ABH and BCH are equal, you should pay attention to the second sign of equality of triangles.
According to the second sign of equality, triangles are equal if a side and two adjacent angles of one triangle are equal to a side and two adjacent corners of the second triangle.
Now consider the triangles ABH and BCH.
1.) BH is the common side of triangles ABH and BCH.
2.) Now it is necessary to prove that the angles that adjoin the side BH are equal.
Consider the angles ABH and CBH at the side BH.
By condition, BH is the height of an isosceles triangle ABC.
By the theorem, the height of an isosceles triangle is also its bisector.
Thus, the angle ABH is equal to the angle CBH.
3.) Now consider the angles AHB and CHB at the side BH.
Since BH is perpendicular to AC, the angles AHB and CHB are 90 °.
Hence, AHB = CHB.
Let us conclude: since BH is the common side of the triangles under consideration, angle ABH = angle CBH, and angle AHB = CHB, then triangles ABH and BCH are equal.