Proving a geometric inequality $BE+CF>EF$

Question

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!

Answer 1

Rotate the figure by $180^\circ$ around $D$.

Now the desired inequality becomes $BE + BF' > EF$.

However $EF = EF'$ by congruent triangles $\triangle EDF' \cong \triangle EDF$.

Finish off using triangle inequality.

Answer 2

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 $GE + GF \gt EF \implies BE + CF \gt EF$.