Assume $ABC$ is not isosceles. Let $M$ be the midpoint of $BC$ and $R$ be the midpoint of arc $\overarc{PQ}$, lying on segment $CM$. Reflect $B$ across $CM$ to get $B' \ne A$. Since $MA=MB=MB'$, $\angle AB'B = 90^{\circ}$ and $AB' \parallel CR$. Also, $\angle CAR = \angle PAR = \angle QBR = \angle CBR = \angle CB'R$. So $AB'$ and $CR$ are the bases of isosceles trapezoid $AB'CR$, and $AR=CB'=CB$ and $AC=B'R=BR$. Thus, $\triangle ABC \cong \triangle BAR$. But this is impossible because $R$ lies inside $\triangle ABC$. Therefore, $ABC$ must be isosceles, specifically $CA=CB$.