In triangle ABC it is known that angle A = 36 degrees, angle B = 72 degrees. The heights AE and BF of the triangle

In triangle ABC it is known that angle A = 36 degrees, angle B = 72 degrees. The heights AE and BF of the triangle intersect at point H. Find the angles of the quadrangle: 1) CFHE 2) ACBH

1) Angle E and angle F are 90 degrees since in a triangle, these are heights.
We find the angle С from this triangle, knowing that the sum of the angles in the triangle is 180 degrees. С = 180- (72 + 36) = 72 degrees. Since the sum of the angles in any four gon is 360 degrees, we find the angle H. H = 360- (90 + 90 + 72) = 108
2) Angle C is 72.
Knowing that in a triangle ABC two angles are equal, then the triangle isosceles sides Ab and AC are equal. So the drawn height AE is also entered as bisector and height. And if it is a bisector, then it divides the angle in half, so the angle A in the ACBH figure is 36: 2 = 18 degrees.
Find the angle H in the triangle AHF 180-90-18 = 72 degrees.
So the angle BHE is 72 because corners are vertical.
This means that the angle H in the ACBH figure is 72 + 72 + 108 = 256 degrees.
Angle B is equal to 18 of the triangle BEC.