In a square, similarity of triangles obtained by joining a vertex to a midpoint

Question

The problem states: $ABCD$ is a square with side $a$. $E$ and $F$ are considered midpoints of sides $AD$ and $AB$ respectively. Let $G$ be the point of intersection of the $EC$ and $DF$ segments. Show that the $\triangle EGD$ and $\triangle DCE$ are similar and deduce that $DG\perp EC$.

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I've tried to prove that $\triangle EDG~\triangle DCE$ using the $AA$ postulate. I made $\angle ECD=α$, then $\angle CED=90°-α$. After that, I used angle properties like alternate interior angles and others all over the figure in the hopes of getting $\angle EDG$ to equal $90°$ in order to show perpendicularity. However, I was unsuccessful in all my attempts. Any idea on how I could solve this?

P.D.: I don't want a solution, just some guidance to solve this myself.

Answer 1

Triangles $DCE$ and $ADF$ are congruent by side-angle-side, showing $\angle EDG = \angle DCE$ and results follow.

Answer 2

An other hint that shows directly the perpendicularity. Let $O$ be the (symmetry) center of the square, and consider a $90^\circ$-rotation centered in $O$, which brings $C\to D$, $D\to A$, $A\to B$, $B\to C$, thus also the mid points of segments one to another, $E\to F$. The lines $CE$ and $EF$ are thus mapped one into the other one by a $90^\circ$ rotation, so they are perpendicular. (These are a lot of sentences, but the idea is optically seen at a glance.)