The distance from the intersection of the middle lines of the two waists of an isosceles triangle to the two waists is equal.

In isosceles △ABC, point D is the intersection of BE and CF, which proves that the distance between point D and AB and AC is equal.

Prove:

In △BCE and △CBF, CE = (1/2) AC = (1/2) AB = BF, ∠BCE = ∠CBF, BC is the same side,

So, △ BCE △ CBF

Available: ∠CBE = ∠BCF,

So, BD = CD.

Connect ads.

In δ△ABD and δ△ACD, AB = AC, BD = CD, and AD are common edges.

So, △ Abd △ ACD,

Available: ∠BAD = ∠CAD,

That is, point D is on the bisector of ∠ABC,

So the distance from point D to AB and AC is equal.