Extend $BI$ to hit the circumcircle at a point $M$. Note that $\frac{\overarc{BP}+\overarc{AQ}+\overarc{AM}}{2}=\frac{\angle A+\angle C+\angle B}{2}=\frac{180}{2}=90$.So $PQ\perp BI$, as desired. QED

Dear Mathlinkers, a consequence of the A, C-Mention's circle... Sincerely Jean-Louis

Attention. This stunt is performed by professional stoopid (i.e. me). Let $R$ be the midpoint of arc $AC$. We claim that $QR^2+PB^2=PR^2+QB^2$. Using Law of Sines for each length, we need to show that $$\sin^2{(\alpha)}+\sin^2{(\beta+\gamma)}=\sin^2{(\beta)}+\sin^2{(\alpha+\gamma)},$$ where $\alpha=\angle ACI$, $\beta=\angle BAI$ and $\gamma=\angle ABI$, thus $\alpha+\beta+\gamma=90^\circ$. Note that for any $x$, we have $\sin{(90^\circ-x)}=\cos{x}$, but since we have, $$\sin^2{\alpha}+\cos^2{\alpha}=1=\sin^2{\beta}+\cos^2{\beta},$$ we are done.

AQPC is cyclic angle APQ= angle ACQ=C/2 ACQB is cyclic angle ACB= angle BAQ=C/2 ACBP is cyclic, angle PAB= angle BCP=A/2 AQPC is cyclic, so angle QPC=180°-angle QAC = 180°-(A+C/2)=B+C/2 angle APC= angle QPC- angle APQ =B+C/2-C/2=B IN ∆ IPC, angle PIC= 180°-( angle ICP+ angle IPC) =90°-B/2 so, angle BIP= angle BIC- angle IPC= 180°-{(B+C)/2}-(90°-B/2)=90°-C/2 Let QP intersect BI at O angle IOP =90°-C/2+C/2=90° So, PQ is perpendicular to BI