The heights of an acute-angled triangle ABC meet at point H. It is known that AB = CH. Find the angle ACB

Consider right-angled triangles CHM and ABM. By condition, AB = CH, then the hypotenuses of the triangles are the same.

Triangles ВKН and СНK are rectangular, in which the angle ВНM = CHM as vertical, then the triangles are similar in acute angle, which means the angle HBK = НСM.

Then the angle ABM = HCM, and the triangle ABM is equal to the triangle CHM in hypotenuse and acute angle.

Then AM = HM, and the right-angled triangle AHM is isosceles, which means the angle MAН = AHM = 45. The triangle AРС is rectangular, in which the angle PA = HAM = 45, then the angle ACP = 180 – 90 – 45 = 45, and therefore the angle ACB = 45.

Answer: Angle ACB is 45.