Circumcenter of acute-$\Delta ABC$ is $O$. Line $AC$ intersects circumcircle of $\Delta AOB$ at point $X$, in addition to vertex $A$. Prove $XO⟂BC$.

Question

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?

Answer 1

$X'$ is the point where $OP$ meets $AC$ (and $OX' \perp BC$)

$2\angle ACB = \angle AOB$ (angle at center is twice the circumference)

but $\Delta AQO \cong \Delta BQO$ by $SAS$ $\implies \angle AOQ = \angle BOQ$ $\angle AOQ + \angle BOQ = \angle AOB \implies \angle BOQ = \angle ACB$

Hence $\Delta CPX' \sim \Delta OQB$ by $AA$ $$\implies \angle AX'O = \angle ABO$$

By the converse of angles on the same segment $X'$ lies on the circumcircle of $\Delta ABO \implies X' = X$