National problem · Geometry

In a rectangle \(ABCD\) with \(|AB| > |BC|\), the perpendicular bisector of the diagonal \(AC\) meets the side \(CD\) at the point \(E\). The circle with center \(E\) and radius \(|AE|\) meets the side \(AB\) at a second point \(F\). Let \(G\) be the foot of the perpendicular dropped from \(C\) to the line \(EF\).

Prove that \(G\) lies on the diagonal \(BD\).

The perpendicular bisector of \(AC\) (short dashes) cuts \(CD\) at \(E\); the circle centered at \(E\) through \(A\) (dotted) returns to \(AB\) at \(F\); the perpendicular from \(C\) to \(EF\) has foot \(G\). The claim is that the diagonal \(BD\) (long dashes) passes through \(G\).