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Geometry Difficulty 7.9 National olympiad, round 2 Prove it Turkey

In a right triangle $ABC$ with $\angle ACB = 90^\circ$ let $D$ be the foot of the altitude from $C$ . Let $E$ and $F$ be the reflections of $D$ with respect to $AC$ and $BC$ , respectively. Let $O_1$ and $O_2$ be the circumcenters of the triangles $ECB$ and $FCA$ , respectively. Prove that
$2|O_1O_2| = |AB|.$

Solution

Let $\angle ABC = \alpha$ and $\angle BAC = 90 - \alpha = \beta$ . $\Rightarrow \angle BCD = \angle EAC = \beta$ and $\angle ACD = \angle ACE = \alpha$ . Let $K$ be the second intersection point of the circumcircle of the triangle $ECB$ and the line $AB$ . $\Rightarrow \angle AKE = \angle KEA = \beta$ and hence $AKE$ is an isosceles triangle.

Note that $O_1$ is the intersection point of the perpendicular bisectors of the line segments $[EK]$ and $[BC]$ . Let $O$ and $L$ be the midpoints of $[AB]$ and $[BC]$ , respectively. Then, $L$ , $O$

and $O_1$ are collinear. By angle chasing we see that $AO_1LC$ is a rectangle. Moreover, since $OLB \cong OO_1A$ we have $|OL| = |OO_1|$ . Consequently, we obtain $OO_1 \parallel AC$ and $|OO_1| = |AC|/2$ . Similarly, one can get $OO_2 \parallel BC$ and $|OO_2| = |BC|/2$ . Therefore, we see that $ABC \sim O_1O_2O$ and hence $|O_1O_2| = |AB|/2$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.