$\angle CDA = \angle CBA = \angle BAC = \angle EAC$ gives us $\triangle CDA \sim \triangle CAE$. Thus, $CB^2 = CA^2 = CD \cdot CE$, and $BC$ is tangent to $(BDE)$.

$ABCD$ is a cyclic quadrilateral, so $\angle CDB=180-\angle BAC$ Because $\triangle ABC$ is isosceles then $\angle EBC=180-\angle CBA=180-\angle BAC$ Thus $\triangle BCE$~$\triangle DCB$ So $\angle DBC = \angle CEB$ By alternate segment theorem, $BC$ is tangent to $(BDE)$

good problem

Observe that $$\angle CBD = \angle CAD = \angle CAB - \angle DAB = \angle CBA - \angle DAB = \angle CBA - \angle DCB = \angle DEB,$$so we are done by Alternate Segment Theorem. $\square$

Solution 2 (Rijul Saini): Inversion about $C$ with radius $CB$ flips $D$ and $E$, so we must have $CD \cdot CE = CB^2$, implying the result by Power of a Point. $\square$ Comment: Oh well I really have to practice using inversion...

Use complex numbers with $a,b,c,d$ on the unit circle with: $c=1$, $b=\overline{a} = \frac{1}{a}$. We can now compute $e$ as: \[ e = \frac{ab(c+d) - cd(a+b)}{ab-cd} = \frac{1+d - da - \frac{d}{a}}{1-d} = \frac{a + ad - da^2 - d}{a(1-d)} \] Now we claim: \[ \vert c-d \vert \cdot \vert c - e \vert = \vert c - b \vert ^2 \]Indeed: \[ \vert (1-d)(1-e) \vert^2 = \left\vert (1-d)\frac{a - ad - a - ad + da^2 + d}{a(1-d)} \right\vert^2 = \left\vert \frac{d(a-1)^2}{a} \right\vert^2 = \frac{d(a-1)^2}{a}\cdot \overline{\left(\frac{d(a-1)^2}{a}\right)} = \frac{d(a-1)^2 \cdot \frac{1}{d}(1-\frac{1}{a})^2}{a \cdot \frac{1}{a}} = \left(\left(1-\frac{1}{a} \right)(1-a)\right)^2\]\[ \vert c-b \vert ^4 = \left(\left(1-\frac{1}{a}\right)(1-a)\right)^2\] Thus our claim holds and we are done by power of a point in $C$.