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

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?

• How does showing that $\angle BOP = \angle BAC$ show that $ABOX'$ is cyclic? It would be better to show that $\angle AX'O = \angle ABO$ and then use converse of angles in the same segment to show that $X'$ lies on the circumcircle of $AOB$ May 7, 2022 at 15:38
• Because when BOP=BAC ,X'OB=180 -BAC which means ABOX' is cyclic May 7, 2022 at 15:50
• That line of reasoning only works when $X'$ lies on the line segment $AC$ but $ABC$ is an acute triangle so this can't happen May 7, 2022 at 16:01
• Are you familiar with barycentric coordinates? They can work pretty well for this problem. May 15, 2022 at 13:28 $$X'$$ 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 $$\Delta ABO \implies X' = X$$