National problem · Geometry

Let \(D\) be the midpoint of side \(AB\) of an acute triangle \(ABC\). Points \(A'\) and \(B'\) are chosen on the segments \(AC\) and \(BC\) so that the triangles \(ADA'\) and \(DBB'\) are both isosceles with apex \(D\) (that is, \(|DA'| = |DA|\) and \(|DB'| = |DB|\)).

Prove that if the line \(CD\) is perpendicular to the line \(A'B'\), then the triangle \(ABC\) is isosceles.

The configuration in a general acute triangle: the four tick marks record \(|DA'| = |DA|\) and \(|DB'| = |DB|\). Here \(CD\) and \(A'B'\) are not perpendicular; the problem asks what follows when they are.