Proving a geometric inequality $BE+CF>EF$ In the figure $D$ is the midpoint of $BC$ and $\Delta EDF$ is right triangle.Prove that $BE+CF>EF$

I tried triangle inequality $$BE+ED>BD$$
$$CD+FC>DF$$
Add:$$BE+FC>DF-ED$$
I am stuck now!


Draw $\triangle DEG \cong \triangle DEB$ ($DG = DB, EG = BE$)
Then $\angle GDF = 90^0 - \angle EDG = 90^0 - \angle BDE = \angle FDC$
Also, $ DG = DB = DC$.
So $\triangle DFG \cong \triangle DFC$ (by S-A-S, $DF$ is common).
So $GE + GF \gt EF \implies BE + CF \gt EF$.
