Let the lines $PQ$ and $AB$ intersect at the point $T$. Then the point $T$ lies on the polars of the points $X,Y$ relative to $(ABC)$, from where $XY$ is the polar $T$ relative to $(ABC)$ and then $(A,B;C',T)=-1$, from where the required follows.

Let $AB,PQ$ meet at $D$ and let $AQ,BP$ meet at $E$ and let $AP,BQ$ meet at $F$. It's well-known that $XY$ passes through $E,F$ so $(AB,C'D) = -1$ so $\frac{AC'}{BC'} = \frac{AD}{BD}$ now using menelaus on $ABC$ and line $A'B'$ we would have $\frac{AC'}{BC'}.\frac{BA'}{CA'}.\frac{CB'}{AB'} = 1$ which proves that $AA',BB',CC'$ are concurrent.