PROBLEM
The circumcenter of an acute-$\mathtt{\Delta ABC}$ is $\mathtt{O}$. Line $\mathtt{AC}$ intersects the circumcircle of the $\mathtt{\Delta AOB}$ at a point $\mathtt{X}$, in addition to the vertex $\mathtt{A}$. Prove that the line $\mathtt{XO}$ is perpendicular to the line $\mathtt{BC}.$
MY APPROACH
- I claim that $\angle ABC$ and $\angle ACB$ can't be equal because then there won't be two points as $X$ and $A$ that intersect the circumcircle of $\Delta AOB$.
- Construct the perpendicular bisector of side $BC$, it is clear that this goes through $O$.
- Let the point at which it intersect $BC$ be $P$ and the point at which it intersect $AC$ be $X'$.
- We know that $\angle BAC=2\cdot\angle BOC$ .The $\Delta BOP$ and $\Delta COP$ are congruent. Therefore we know that $\angle BOP=\angle BAC$, and $\angle X'OB=180^o -\angle BAC$.
- Thus, we conclude that $ABOX'$ is a cyclic quadrilateral and $X'$ also lies on the circumcircle of $\Delta AOB$. Therefore, $X=X'$, which means $OX$ is perpendicular to $BC$.
Is my Proof Correct?