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 $\mathtt{BC}.$


  • 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?

  • $\begingroup$ 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 $AOB$ $\endgroup$
    – Arizus
    May 7, 2022 at 15:38
  • $\begingroup$ Because when BOP=BAC ,X'OB=180 -BAC which means ABOX' is cyclic $\endgroup$
    – SC 16
    May 7, 2022 at 15:50
  • $\begingroup$ 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 $\endgroup$
    – Arizus
    May 7, 2022 at 16:01
  • $\begingroup$ Are you familiar with barycentric coordinates? They can work pretty well for this problem. $\endgroup$
    – ZNatox
    May 15, 2022 at 13:28

1 Answer 1


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$X'$ 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 $\Delta ABO \implies X' = X$


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