If $S$ is the circumcenter of $\triangle ABC$ and $D, E, F$ are the feet of altitiudes to opposite sides from $A, B, C$, prove that $SB$ is perpendicular to $DF$.
Here $\angle BAD = \angle BCF$, $\angle EBC = \angle DAC$ and $\angle ABE = \angle ACF$.
I've tried doing this by trying to create congruent triangles with bases $OB$ and $OC$ (I've taken $O$ as the midpoint of $BC$.) but have not made any progress. Please help.
Source: Challenge and Thrill of Pre-College Mathematics
On
$\angle 7 = \angle 1 + \angle 5 = \angle 2 + \angle 5$ [because there is a red dotted circle].
… = $\angle 3 + \angle 5$ [because there is a blue dotted circle].
… = $\angle4 + \angle 5$ [because of the black circle].
… = $\angle 4 + \angle 6$ [because of the black circle]
… = $90^0$ [angle in semi-circle]
On
The angle $\widehat{BSC}$ equals $2\widehat{A}$ since $S$ is the centre of the circumcircle of $ABC$, hence the angle between $BS$ and the perpendicular bisector of $BC$ equals $\widehat{A}$. $ACDF$ is a cyclic quadrilateral, hence the angle between $FD$ and $BC$ equals $\widehat{A}$, too, and the claim easily follows.
Define $K$ as midpoint of $BC$. Call the intersection of $BS$ and $DF$ as $P$. Since $S$ is the center of the circle of $ABC$, we have $\angle BSC=2\angle A$. Since $SB=SC$, SK bisects $\angle BDF$ therefore $\angle BSK=\angle A$. On the other hand since $D$ and $F$ are feet of the altitudes $CAFD$ is cyclic, hence $\angle BDF = \angle A$. Now $\angle A = \angle BSK = \angle BDF$ tells us $KSPD$ is a cyclic quadrilateral. Hence $\angle SKB=\angle DPS=90^{\circ}$.