$BO$ cut $(O)$ again at $Y.$ $AC$ is Simson line with pole $Y$ and $EF$ is Simson line with the pole $D$ $\Longrightarrow$ $\angle CEF = \frac{_1}{^2} \angle DOY = \angle DBY.$ Combined with $BED \perp AC$ $\Longrightarrow$ $BOY \perp EF.$

$CDEF$ is cyclic and $\angle EDF=C, \angle BDC=A\Longrightarrow \angle FEC=\angle FDC=A-C$. The angle between the tangent at $B$, $\omega$ and $BE$ is $90^{\circ}-A+C=\angle BEF$ and so, $EF\parallel \omega\perp BO$. Alternate proof: Let $BO$ intersect $EF$ at $K$. $DCFE$ is cyclic and $\angle BDC=A\Longrightarrow \angle EFD=\angle ECD=90^{\circ}-A$ and so, $\angle EFB=A$. We also have $\angle KBF=\angle ABE$ and so, $\angle FBK\sim \angle ABE$ and hence, $BO\perp EF$

Dear Mathlinkers, according to Heine's theorem, the tangent at the circumcircle at B give the direction of the Simson's line with pole D, and we are done. Sincerely Jean-Louis

Proof 1. $m\left(\widehat{OBF}\right) +m\left(\widehat{BFE}\right)=$ $m\left(\widehat{ABD}\right)+m\left(\widehat{CDE}\right)=$ $m\left(\widehat{DCE}\right)+m\left(\widehat{CDE}\right)=$ $90^{\circ}\implies$ $BO\perp EF$ .

Another Proof: Let $OB$ cuts $EF$ at $K$ $(ACB)=\alpha,(FDC)=\beta,(EBO)=\theta$ then $(EKB)=90 \Longleftrightarrow \beta=\theta$ (EDFC is quadrilateral beams) then using $O$ is circumcenter we have $(ABO)=90-(ACB) $ $\Longleftrightarrow 90-\alpha-\beta+\theta=90-\alpha \Longleftrightarrow \beta=\theta$ Done!

PP. The line passing through $B$ is perpendicular to the side $AC$ at $E$ . This line meets the circumcircle of $\triangle ABC$ at $D$ . The foot of the perpendicular from $D$ to the side $BC$ is $F$ . If $O$ is the center of the circumcircle of $\triangle ABC$ , then prove that $BO\perp EF$ . Proof 1. $m\left(\widehat{OBF}\right) +m\left(\widehat{BFE}\right)=$ $m\left(\widehat{ABD}\right)+m\left(\widehat{CDE}\right)=$ $m\left(\widehat{DCE}\right)+m\left(\widehat{CDE}\right)=$ $90^{\circ}\implies$ $BO\perp EF$ . Proof 2. The lines $BB^{(*)}$ and $EF$ are antiparallel lines to $DC$ in $\triangle BDC\implies$ $EF\parallel BB\perp OB\implies$ $OB\perp EF$ . $(*)$ I denoted $BB$ - the tangent line in $B$ to the circumcircle of $\triangle ABC$ .

1) $\angle FBO=\angle CBO=90^\circ-\angle BAC$ 2) $ABCD, CDEF$ are cyclic quads, hence $\angle EFB=\angle BDC=\angle BAC$, so $\angle EFB+\angle OBF=90^\circ$, done. Best regards, sunken rock

Inversion one-liner. With inversion $(B; \sqrt{BE\cdot BD})$ we have $EF\mapsto (ABC)$ and clearly $BO\perp (ABC)$.

Dear Mathlinkers, 1. (X) the circle with diameter DC 2 Reim's theorem to (X) and (O) : EF // to the tangent to (O) at B and we are done... Sincerely Jean-Louis