Since $\angle BAD = 90^\circ$ (Fig. 1), $BD$ is the diameter of circle $c$ and therefore $CD = BD$. Since $BD$ is diameter, also $\angle BED = 90^\circ$, so $DE$ is an altitude of the isosceles triangle $BDC$, bisecting its base $BC$. Hence $E$ is the midpoint of the hypotenuse $BC$ of the triangle $ABC$. Since the midpoint of the hypotenuse is the circumcentre of a right triangle, it follows $EA = EB$. This means that $ABE$ is an isosceles triangle.

Fig. 1

As in the previous solution, we show that $CD = BD$. Hence $\angle ECD = \angle EBD$. From the equality of the inscribed angles subtending the arc ED it also follows $\angle EBD = \angle EAD$. From the triangle ABC we get $\angle ABC = 90^\circ - \angle BCA$, or $\angle EBA = 90^\circ - \angle ECD$. On the other hand, $\angle EAB = \angle DAB - \angle EAD = 90^\circ - \angle EBD = 90^\circ - \angle ECD$. Consequently $\angle EBA = \angle EAB$. So the triangle ABC is isosceles.