Prove that if two heights of a triangle are equal, then this triangle is isosceles.

Let a triangle ABC be given in which two heights AP and CK are drawn, with AP = CK. Consider right-angled triangles APC and CKA (they have ∠APC = 90 ° and ∠CKA = 90 °).
In these triangles:
1) the vertex B is common, therefore, ∠BAP = ∠BCK = 90 ° – ∠B;
2) AP = СK – by condition.
It turns out that ∆APC = ∆ CKA according to the 2nd sign of equality of right-angled triangles (along the leg and the opposite acute angle), then their hypotenuses will also be equal to AB = BC. And these are the sides of ∆ ABC, so it is isosceles.
From this we can conclude that if the two heights of a triangle are equal to each other, then this triangle is isosceles.
Q.E.D.