It is obvious that $\angle XBC=\angle BCY=90^\circ \rightarrow BX \parallel CY$. Let the intersection of line $OC$ and $BX$ be $T$. Then $T$ lies on the circumcircle of $\triangle ABC$, so it follows that $OC=OT$. So $O \text{ is midpoint of } XY \longleftrightarrow XT=CY \longleftrightarrow \frac{c}{\text{sin} B} - 2RcosA=\frac{a}{\text{tan} B} \longleftrightarrow \text{sin} C=\text{cos}A\text{sin}B+\text{sin}A\text{cos}B$ with our usual notations. (The last equivalence came from the sine law.)

omar31415 wrote: Let $ABC$ an acute triangle with circumcenter $O$. Points $X$ and $Y$ are chosen such that: $\angle XAB = \angle YCB = 90^\circ$ $\angle ABC = \angle BXA = \angle BYC$ $X$ and $C$ are in different half-planes with respect to $AB$ $Y$ and $A$ are in different half-planes with respect to $BC$ Prove that $O$ is the midpoint of $XY$. Let $B'$ be the antipode of $B$. Then, using angle chasing we obtain $BXB'Y$ is a parallelogram.So we're done!

Setting $A=(a,b)\,\,B=(0,0)\,\,C=(c,0)$ so $X$ is the intersection of the line perpendicular to $AB$ and y-axis. Solving the equation $y=-\frac{a}{b}x+\frac{a^2+b^2}{b}$ and $x=0$ gives the coordinates of $X(0,\frac{a^2+b^2}{b})$. Similarly we get the coordinates of $Y(c,-\frac{ac}{b})$. Now solving the equations of perpendicular bisector of $AB$ and $BC$ gives the coordinates of $O$ as $(\frac{c}{2},\frac{a^2-ac+b^2}{2b})$ which is clearly the midpoint of $XY$.

Its clear that $XA$ meets $CY$ at the antipode of $B$ on $(ABC)$ so lets call it $B'$. The angles condition can be re-writen as $(BAX)$ tangent to $BC$ and $(BYC)$ tangent to $AB$ hence we get that $\angle XAB=\angle XBC=90=\angle YBA=\angle YCB$ which means that $XB \parallel YB'$ and $YB \parallel XB'$ hence $XBYB'$ is a paralelogram thus we are done