13. Let $A B = c, A C = b, \angle C B A = \beta, B C = a$, and $A D = h$. Let $r_{1}$ and $r_{2}$ be the inradii of $A B D$ and $A D C$ respectively and $O_{1}$ and $O_{2}$ the centers of the respective incircles. We obviously have $\frac{r_{1}}{r_{2}} = \frac{c}{b}$. We also have $D O_{1} = \sqrt{2} r_{1}$, $D O_{2} = \sqrt{2} r_{2}$, and $\angle O_{1} D A = \angle O_{2} D A = 45^{\circ}$. Hence $\angle O_{1} D O_{2} = 90^{\circ}$ and $\frac{D O_{1}}{D O_{2}} = \frac{c}{b}$ from which it follows that $\triangle O_{1} D O_{2} \sim \triangle B A C$. ! We now define $P$ as the intersection of the circumcircle of $\triangle O_{1} D O_{2}$ with $D A$. From the above similarity we have $\angle D P O_{2} = \angle D O_{1} O_{2} = \beta = \angle D A C$. It follows that $P O_{2} \parallel A C$ and from $\angle O_{1} P O_{2} = 90^{\circ}$ it also follows that $P O_{1} \parallel A B$. We also have $\angle P O_{1} O_{2} = \angle P O_{2} O_{1} = 45^{\circ}$; hence $\angle L K A = \angle K L A = 45^{\circ}$, and thus $A K = A L$. From $\angle O_{1} K A = \angle O_{1} D A = 45^{\circ}, O_{1} A = O_{1} A$, and $\angle O_{1} K A = \angle O_{1} D A$ we have $\triangle O_{1} K A \cong \triangle O_{1} D A$ and hence $A L = A K = A D = h$. Thus
$$\frac{E}{E_{1}} = \frac{a h / 2}{h^{2} / 2} = \frac{a}{h} = \frac{a^{2}}{a h} = \frac{b^{2} + c^{2}}{b c} \geq 2.$$
Remark. It holds that for an arbitrary triangle $A B C, A K = A L$ if and only if $A B = A C$ or $\measuredangle B A C = 90^{\circ}$.