In an isosceles triangle ABC with a base AC, the mid-perpendicular of side AB intersects

In an isosceles triangle ABC with a base AC, the mid-perpendicular of side AB intersects the base of AC at point P. Find angle C if ABP is 52 °.

Since the segment РН is the midpoint perpendicular of the triangle ABC, then the segment AH = BH, and the triangles AРН and BPН are rectangular.

In right-angled triangles AВН and ВРН, the leg РН is common and legs AB and BН are equal, then the right-angled triangles AВН and ВРН are equal in two legs, which means the angle РВН = РAН = 52.

Since triangle ABC is isosceles, the angle ACB = CAB = РAН = 52.

Answer: The ACB angle is 52.