The heights of an isosceles triangle, drawn from the apex at the base, at the intersection form an angle

The heights of an isosceles triangle, drawn from the apex at the base, at the intersection form an angle of 140 degrees, find the angle at the base

The AOH angle is adjacent to the AOB angle, the sum of which is 180, then the AOH angle = (180 – AOB) = (180 – 140) = 40.

Triangle AOН is rectangular, in which the angle OHA = 90, the angle AOH = 40, then the angle OAН = (90 – 40) = 50.

The AMN triangle is rectangular with a right angle AMC, then the angle AFM = (90 – CAM) = (90 – 50) = 40.

Then the angle CAB = CBA = (180 – ABC) / 2 = (180 – 40) / 2 = 70.

Answer: The angle at the base is 70.