Let $E$, $F$ and $G$ be the centers of excircles touching $BD$, $AD$ and $AB$ respectively, and let $\omega$ be the circumcircle of $ABC$ (see Fig. 5). To prove the assertion of the problem, we will show that $F$ and $G$ cannot both lie on $\omega$ and that $E \in \omega \iff |BD| = |DC| \iff F \in \omega$.

As $AF$ and $AG$ are bisectors of the two complementary angles of $BAD$, the point $A$ lies on the segment $FG$. Thus, only one of the rays $AF$ and $AG$ can cut the circle $\omega$ again, and therefore only one of $F$ and $G$ can lie on $\omega$.

To show that $E \in \omega$ iff $|BD| = |DC|$, we first note that $E$ lies on $\omega$ iff $\angle CAE = \angle CBE$. Since $\angle CAE = \frac{1}{2}\angle BAD = \frac{1}{2}(\pi - \angle ADB - \angle ABD) = \frac{1}{2}(\pi - (\pi - 2\angle BDE) - (\pi - 2\angle DBE)) = \angle BDE + \angle DBE - \frac{\pi}{2} = \frac{\pi}{2} - \angle BED$, we see that $E \in \omega$ is equivalent to $\angle CBE = \frac{\pi}{2} - \angle BED$, i.e $BC$ and $DE$ being perpendicular. Since $DE$ is the bisector of $BDC$, this occurs iff $|BD| = |DC|$.

Fig. 5

It remains to show that $|BD| = |DC|$ iff $F \in \omega$. Point $F$ lies on $\omega$ iff $\angle AFB = \angle BCD$. Using the fact that $\angle BAF = \angle BAD + \frac{1}{2}(\pi - \angle BAD) = \frac{1}{2}(\pi + \angle BAD)$, we get $\angle AFB = \pi - \angle ABF - \angle BAF = \frac{\pi - \angle ABD - \angle BAD}{2} = \frac{\angle ADB}{2} = \frac{\angle BCD + \angle CBD}{2}$. Thus $\angle AFB = \angle BCD$ is equivalent to $\angle BCD = \angle CBD$, i.e., $|BD| = |DC|$.