Let's denote the radius of the circumscribed circle around the $\triangle ACD$ as $R_1$, the radius of the circumscribed circle around the $\triangle BDC$ as $R_2$. So, if $CD = l$ then by the law of sines (Fig.47):
$$\frac{l}{\sin \beta} = 2R_1 \Rightarrow \sin \beta = \frac{l}{2R_1}.$$
Herewith, $\cos \varphi = \frac{DE}{DQ} = \frac{l}{2R_1} = \sin \beta$.
Therefore, from the acute triangle $\triangle ABC$ $\varphi = \frac{\pi}{2} - \beta$, and similarly $\psi = \frac{\pi}{2} - \alpha$.
Therefore $\angle PDQ = \varphi + \psi = \frac{\pi}{2} - \alpha + \frac{\pi}{2} - \beta = \pi - \alpha - \beta = \gamma = \angle ACB$.

Fig. 47

Moreover, from the following equation:
$$\frac{QD}{PD} = \frac{R_1}{R_2} = \frac{l}{2\sin\beta} \cdot \frac{2\sin\alpha}{l} = \frac{\sin\alpha}{\sin\beta} = \frac{a}{b} = \frac{CB}{AC}$$

the sides about the equal angles of our triangles are proportional, so triangles $ABC$ and $DPQ$ are similar for any point $D \in AB$. Points $A, B$ do not satisfy the condition, otherwise one of the $\triangle ACD$ or $\triangle BDC$ degenerates to a line segment.