Problem 1310

Let $E$ denote the intersection of the lines $AD$ and $BC$, let $F$ denote the intersection of $BD$ and $CA$ and let $G$ denote the intersection of $CD$ and $AB$. The equality $\angle BAD = \angle DCB$ implies that the triangles $GAD$ and $ECD$ are similar since they have two common angles. So,
$$\frac{|GD|}{|AD|} = \frac{|ED|}{|CD|}.$$
Similarly, the equality $\angle CBD = \angle DAC$ implies that the triangles $FAD$ and $EBD$ are similar, so

$$\frac{|AD|}{|FD|} = \frac{|BD|}{|ED|}.$$
If we multiply the above inequalities we get
$$\frac{|GD|}{|FD|} = \frac{|BD|}{|CD|} \quad \text{or} \quad \frac{|GD|}{|BD|} = \frac{|FD|}{|CD|}.$$
Since $\angle GDB = \angle CDF$, we conclude that the triangles $GDB$ and $FDC$ are similar, so $\angle DBG = \angle FCD$ and $\angle DBA = \angle ACD$. This and the assumptions of the problem imply that
$$\angle BAD + \angle CBD + \angle DBA = \frac{1}{2}(\angle BAC + \angle ACB + \angle CBA) = 90^\circ,$$
so $\angle AEB = 180^\circ - (\angle BAD + \angle CBD + \angle DBA) = 90^\circ$, which is what we wanted to show.