Proved as follows:

As shown in the figure below, in isosceles Δ AB, AC, CD and Be are the heights on the sides of AB and AC, respectively, and intersect with F.

Because Δ ABC is an isosceles triangle, BA=AC, ∠ABC=∠ACB.

In rtδADC and rtδAEB, there are ∠DAC=∠EAB and AB=AC.

So rt δ ADC ≌ rt δ AEB

So ∠Abe =∞-①.

Because again, ∠ ABC = ∠ AC B-②.

And ∠ABC =∠ Abe +∠FBC, ∠ACB=∠ACD+∠FCB? - ③

At the same time, 12③, available, ∠FBC=∠FCB.

Therefore, δδBFC is an isosceles triangle.

So FB=FC

That is, the intersection of the heights of the two waists on the side of a triangle is equal to the distance between the two ends of the bottom.

The process is detailed enough, I hope I can help you. If you feel satisfied, set it as a satisfactory answer!