# In a triangle ABC, the height BD divides angle B into two angles, and the angle ABD = 40 degrees

**In a triangle ABC, the height BD divides angle B into two angles, and the angle ABD = 40 degrees, the angle CBD = 10 degrees a) prove that ABC is isosceles, and indicate its base b) the heights of this triangle intersect at point O. Find the angle BOC**

Angle ABC = ABD + CBD = 40 + 10 = 50.

Since BD is the height of triangle ABC, triangles BCD and ABD are rectangular.

Then the angle ВСD = ВСА = (180 – ВDC – CBD) = (180 – 90 – 10) = 80.

Angle BAD = (180 – ADB – ABD) = (180 – 90 – 40) = 50.

Since the angle ABC = BAC = 50, the triangle ABC is isosceles with the base AB.

Since triangle ABC is isosceles, its height CH is also the bisector of angle C, then the angle BCH = BCO = BCA / 2 = 80/2 = 40.

Then in the triangle BOS the angle BOC = (180 – BCO – CBO) = (180 – 40 – 10) = 130.

Answer: AB is the base of the triangle, the angle BOC is 130.