Let $O$: circumcenter, $BE,CF$: altitudes, $AO\cap BC=A_1$ Since $AO\bot EF$, $$AP=AQ \implies AO \bot PQ \implies EF \parallel B_1 C_1 \implies \frac{AB_1}{AC_1} = \frac{AE}{AF}$$$\triangle AEB \sim \triangle AFC \implies \frac{AE}{AF} = \frac{c}{b}$. And $AB_1 = c \cdot \frac{\cos C}{\cos(C-A)}, AC_1 = b \cdot \frac{\cos B}{\cos(B-A)} \implies \frac{AB_1}{AC_1} = \frac{c}{b} \frac{\cos C \cos (B-A)}{\cos B \cos(C-A)}$. Therefore, $$\frac{AB_1}{AC_1} = \frac{AE}{AF} \implies \frac{\cos(B-A)}{\cos B} = \frac{\cos(C-A)}{\cos C} \implies \cos A + \sin A \cot B = \cos A + \sin A \cot C \implies B=C$$

Since $\measuredangle AC_1B_1$ $=$ $\frown (AQ + BP)/2$ $=$ $\frown (AP + BP)/2$ $=$ $\measuredangle ACB$ so $BCB_1C_1$ is cyclic and therefore $\measuredangle BB_1C$ $=$ $\measuredangle BAC$ $+$ $90$ $-$ $\measuredangle ACB$ $=$ $\measuredangle BC_1C$ $=$ $\measuredangle BAC$ $+$ $90$ $-$ $\measuredangle ABC$ hence $\measuredangle ABC$ $=$ $\measuredangle ACB$ and we are done.