Behold! my laser-precise hand-drawn diagram We note that $\angle B$ must be an acute angle, or that $D$ must lie within line segment $BC$. This is because $\angle DAB$ and $\angle DAE$ cannot be equal, since $E$ must lie within line segment $BC$, which makes $E \neq B$. Thus, $D$ lies between $B$ and $C$. Additionally, since $E$ is an angle bisector, and $\angle BAD < \angle DAC$ (since $|BA| < |AC|$), $E$ must be between $D$ and $C$. We also notice that $E$ must be closer to $B$ than $F$, since by the angle bisector theorem, $|BA|<|AC|$ means $|BE|<|EC|$. Therefore, the four equivalent angles at the end of the problem sum up to angle $\angle BAC$, and the points along $BC$ occur in the order shown in the diagram. Let angle $\angle BAD$ in degrees be $x$. Let $AB$ be length $c$, and let $AC$ be length $b$. Let $BD$ be length $p$ and $EF$ be length $q$. Reflect $E$ over $AF$ to get point $E'$ (which lies on $AC$ because $\angle EAF = \angle FAC$). Let the intersection of $EE'$ be $D'$. Because the angle bisector of $\angle BAE$ meets $BE$ at a right angle, $\triangle BAE$ is an isosceles triangle, and so is $\triangle AEE'$, due to the same reasoning. Since $\angle BAE = \angle EAC$ and $AB = AE = AE'$, $\triangle BAE \cong \triangle EAE'$. This means that $|BE| = |EE'| = 2p$. Additionally, since $\triangle AE'F$ is a reflection of $\triangle AEF$ over $AF$, we get that $EF = FE' = q$. From right triangle $\triangle ABD$, we get $\angle ABD = 90^{\circ}-x$, and from that, we use the angles in $\triangle ABC$ to get $\angle ACB = 90^{\circ}-3x$. From $\triangle ADE$, we get $\angle AED = 90^{\circ}-x$. Since $\triangle BAE \cong \triangle EAE'$, $\angle AEE' = \angle AE'E = 90-x$. At point $E$, $\angle E'EC = 180^{\circ} - \angle AEB - \angle AEE' = 2x$, and since $\triangle EE'F$ is isosceles, $\angle EE'F = 2x$. This means $\angle FE'C = 180^{\circ} - \angle AE'E - \angle EE'F = 90^{\circ}-x$. Finally, by triangle $\triangle E'FC$, $\angle E'FC = 4x$. From these angles that we have found, we find that $\triangle ABC \sim \triangle FE'C$ and $\triangle AEC \sim \triangle EE'C$. From the first similarity, we find that $\frac{|AB|}{|AC|} = \frac{|FE'|}{|FC|}, \frac{c}{b} = \frac{q}{2p+q}$. From the second similarity, we find that $\frac{|AE|}{|AC|} = \frac{|EE'|}{|EC|}, \frac{|c|}{|b|} = \frac{2p}{2p+2q} = \frac{p}{p+q}$. Therefore: \[\frac{q}{2p+q} = \frac{p}{p+q} \Rightarrow pq + q^2 = 2p^2 + pq \Rightarrow 2p^2 = q^2\]As such, $\sqrt{2}p = q$, since both values are positive. Using the pythagorean theorem on $\triangle DEF$, we find that $|DE|^2 + p^2 = q^2 = 2p^2 \Rightarrow |DE|^2 = p^2 \Rightarrow |DE| = p$. Thus, $\triangle DEF$ is a right-angled isosceles triangle, and $\angle DEF = 2x = 45^{\circ}$, $x = \frac{45}{2}^{\circ}$. Therefore, $\angle ABC = 90^{\circ} - x = \frac{135}{2}^{\circ}$.