In a triangle, $E$ is the midpoint of $AC$. Let $\angle BCE = \angle ABE$. Prove that $AB + BD = CD$ (where $D$ is the midpoint of $BC$), i.e., $AB + BD = l_1 + l_2$.
In $\triangle BCE$ and $\triangle ABE$: $BE^2 = EC^2 + BC^2 - 2\,EC\cdot BC\cos\angle BCE$ $BE^2 = AE^2 + AB^2 - 2\,AE\cdot AB\cos\angle ABE$ $AB = EC$ because $E$ is the midpoint, and $\angle BCE = \angle ABE$. Subtracting: $BC^2 - AB^2 = 2\,EC\cos\angle BCE(BC - AB)$ $\Rightarrow BC + AB = 2\,EC\cos\angle BCE$ In $ACDE$: $BC + AB = 2CD$, so $BD + CD + AB = 2CD$, hence $BD + AB = l_1 + l_2$. Method 2: Let $A'$ be on $CB$ extended such that $AB = A'B$. Then $EB$ bisects $\angle ABA'$, so $EA = EA' = EC$. $D$ is midpoint of $A'C$. Then $AB + BD = A'B + BD = A'D = \tfrac{1}{2}A'C = DC$.
In a triangle, $E$ is the midpoint of $AC$. Let $\angle BCE = \angle ABE$. Prove that $AB + BD = CD$ (where $D$ is the midpoint of $BC$), i.e., $AB + BD = l_1 + l_2$.