geometry - Problem associated with the use of Intercept theorem for the triangle

In triangle $ABC$ on the side $AC$ point $D$ is chosen so that $BC = BD$. On the side $AB$ selected points $P$ and $Q$ so that $\angle PDA = \angle QCA = \angle BAC$. Need to prove that $AP = BQ$.

I have the following idea. Note that the triangles $APD$ and $AQC$ are isosceles. Then $AP = PD$ and $AQ = QC$. Next apply the Intercept theorem, we obtain

$$
\frac{AP}{AD} = \frac{PQ}{DC}.
$$

I'm stuck at this step. In what direction you need to continue to argue?