Croatian Mathematical Olympiad

Let $O$ be the circumcentre of the triangle $ABC$.



Point $D$ lies on the angle bisector of $\angle BAC$. Therefore $A$, $I$ and $D$ are collinear. Lines $IJ$ and $OD$ are parallel, since they are both perpendicular to $BC$.

We will prove the claim by showing that the triangles $AOI$ and $EOI$ are congruent.

First we show that the triangles $AIO$ and $IJD$ are similar. Note that $\angle IAO=\angle DAO=\angle ODA=\angle ODI=\angle JID$. To show similarity, we need to prove that $|AI|:|AO|=|IJ|:|ID|$, i.e. $$|AI|\cdot |ID|=|AO|\cdot |IJ|.(1)$$ We have $|IJ|=2r$ and $|AO|=R$, where $r$ and $R$ denote the incircle and the circumcircle radii, respectively. Therefore, the right-hand side of (1) equals $2Rr$. The left-hand side product $|AI|\cdot |ID|$, is in fact the power of point $I$ with respect to the circumcircle of the triangle $ABC$ and equals $R^2-|OI{|}^2$. Now it suffices to verify that $R^2-|OI{|}^2=2Rr$, which holds because of Euler's theorem which expresses the distance between the incentre and the circumcentre using this exact formula.

Therefore, we can conclude that the triangles $AIO$ and $IJD$ are similar. This means that $\angle OIA=\angle DJI$, and $\angle DIO=\angle IJE$ follows. Since lines $IJ$ and $OD$ are parallel, we have $\angle IJE=\angle ODE$, and since $|OE|=|OD|$ we have $\angle ODE=\angle DEO$. Now having $\angle DIO=\angle DEO$ we conclude that points $D$, $O$, $I$ and $E$ are concyclic.

From this we have $\angle IOE=\angle IDE$, and from the proven similarity we have $\angle IDE=\angle IDJ=\angle AOI$. It follows that $\angle IOE=\angle AOI$. Since $|AO|=|EO|$ and the side $\overline{OI}$ is mutual, we conclude that the triangles $EOI$ and $AOI$ are congruent.