Croatian Mathematical Competitions

Let $\alpha$, $\beta$, $\gamma$ be the angles of the triangle $ABC$ and $R$ the radius of its circumcircle. Without loss of generality, let $E$ lie on the arc $BC$ and $F$ on the arc $AC$.
Since $\angle CAD = \angle CBD = 90^\circ$, the point $D$ lies on the circumcircle of $ABC$ and $CD$ is the diameter.

Since $|CE| = |CH|$, $\angle BEC = 180^\circ - \alpha = \angle BHC$ and $BC$ is the common side of $BEC$ and $BHC$, these triangles are congruent by "S-S-A" theorem (angle $180^\circ - \alpha$ is obtuse). Thus we have $\angle ECB = \angle HCB$, and the line $BC$ is the axis of symmetry of the isosceles triangle $ECH$. Therefore $BC \perp HE$ and we conclude that the points $E$, $H$ and $A$ lie on the same line.

From the right triangle $CDE$ we have
$$|DE| = |CD| \cos \angle CDE = |CD| \cos \angle CAE = |CD| \cos(90^\circ - \gamma) = |CD| \sin \gamma.$$
By the law of sines we have $2R \sin \gamma = |AB|$ and we get $|DE| = |AB|$.
Analogously we get $|DF| = |AB|$.