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?
On
We have that the perpendicular bisector of $CD$ goes through $B$, the perpendicular bisector of $AD$ goes through $P$ and $PD\parallel QC$. Let $Q'$ be the projection of $Q$ on $AC$: since $AQ=QC$, $Q'$ is the midpoint of $AC$. Let $P',B'$ be the projections of $P,B$ on $AC$. It is enough to show that $AP'=Q'B'$, or $AQ'=P'B'$. On the other hand $AQ'=\frac{1}{2}AC$ and $$ P'B'=P'D+DB' = \frac{1}{2}AD+\frac{1}{2}DC = \frac{1}{2}AC.$$
We can call $\angle PDA = \angle QCA = \angle BAC=\alpha$ and $\angle BDC=\angle BCD=\theta$.
$1)$ Once $\angle PDA = \angle QCA$ we have that, by Tales's Theorem, $QC||PD$ and then $\angle DPB=\angle CQB$
$2)$ We also have that $\angle PDB = 180º-\alpha-\theta$ and $\angle QBC=180º-\alpha-\theta$, so $\angle PBD=\angle BCQ$.
Finaly, using $(1)$, $(2)$ and $BD=BC$ we get that $\Delta DPB\equiv\Delta BQC$ by the case (angle, side, angle), and then $BQ=PD$, but $PD=AP$ because $\Delta APD$ is isosceles (because $\angle PDA = \angle BAC$) and then $AP=BQ$.