Geometry problem (from national competition) Given right angled triangle $\triangle ABC$ with right angle at point C.  
Let points $D, E$ lie on $AB$, such that $|BC|=|BD|$ and $|AC|=|AE|$.  
Let point $F$ be orthogonal projection of point $D$ onto $AC$ and let point $G$ be orthogonal projection of point $E$ onto $BC$.
Prove $|DE|=|DF|+|EG|$.

What I've got so far:
EDIT: Had a picture here, but I removed it because it was incorrect and now I don't feel like making new one because the question is already answered.
How can I solve or approach this problem?

Drop the altitude from $C$ to $AB$ which cuts $AB$ at $X$. Then it is enough to prove that $DX=DF$. Say $\angle ABC = 2x$, then $$ \angle BDC = \angle DCB = 90-x$$ and so $\angle DCX = x$. Clearly we have $$\angle ACX = 90-\angle XCB = 2x,$$ so $\angle FCD = x$.
So triangles $FCD$ and $XDC$ are congurent (a.s.a.) and thus a conclusion.
