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.
$$|AI| \cdot |ID| = |AO| \cdot |IJ|. \qquad (1)$$
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.