Question 53
Let $H$ be the orthocenter of the acute triangle $ABC$. Let $X$ be a point in the plane such that the triangle $XAH$ is an isosceles right triangle, with the hypotenuse $AH$, and the points $B$ and $X$ are separated by the line $AH$. Prove that $\overrightarrow{XA} + \overrightarrow{XC} + \overrightarrow{XH} = \overrightarrow{XB}$ if and only if $\angle BAC = 45^\circ$.