For reference(exact copy of the question): In a triangle with acute angle $ABC$. the straight line of Euler determines with its sides an inscribed quadrilateral. Find the measure of the angle that forms the straight Euler line and the circumradius that passes through B.(answer: $90^o$)
My progress:: I confess that I didn't understand the question well... I made the drawing below but only in function of the answer and I'm not sure if that's what the elucidate is asking... But I can't imagine how to solve it from a generic drawing.
Resolution figure by MathLover:
The question says $AMNC$ is an inscribed quadrilateral, so $\angle AMN = 180^\circ - \angle C$.
$\implies \angle BMO = \angle C$
Also as $O$ is the circumcenter of $\triangle ABC$, $\angle AOB = 2 \angle C$ and as $\triangle AOB$ is isosceles,
$\angle MBO = \angle BAO = \frac{180^\circ - 2 \angle C}{2} = 90^\circ - \angle C$
That leads to $\triangle BOM$ being a right triangle and $\angle BOM = 90^\circ$.
Also note that this is possible when $\angle ABC$ is acute and the Euler line passes through the sides of $\angle ABC$ such that $AMNC$ is cyclic. In other words, $A$ and $C$ should be the vertices of the quadrilateral and not $B$.