Here's an elementary geometry proof. The left picture is the quadrilateral of relevance in your picture. You want to show that given that $AD\cong CE$ and $\angle DAC+\angle ECA<180^\circ$ (since they are two of the three angles of $\triangle ABC$), we have $DE.

Since $\angle DAC+\angle ECA<180^\circ$, we can construct exterior to the quadrilateral $ACDE$ a segment $AF$ (in red) such that $AF$ is congruent and parallel to $CE$. Now ACEF is a parallelogram since it has one pair of congruent and parallel opposite sides, so $AC\cong FE$, and it suffices to show that $DE. Consider the triangle $\triangle DEF$ (in blue). Since the angle opposite the greater side is greater, it suffices to show that $\angle FDE>\angle DFE$. But this is easy since $ {\small \angle FDE=\angle FDA+\angle ADE=\angle AFD+\angle ADE=(\angle AFE+\angle DFE)+\angle ADE,} $ which is clearly greater than $\angle DFE$, as desired. (Note that in the second step, we used that $\angle FDA=\angle AFD$ since $\triangle ADF$ is isosceles.)