Macedonian Mathematical Competitions

$AB$ intersects the circumference in a point $M$. The line, which passes through $M$ and is parallel to $CD$, intersects the circumference in a point $N$. Because $\angle MDC = 90^\circ$, the segment $MC$ passes through the centre of the circle, which means that $\angle MNC = 90^\circ$, we conclude that the quadrangle $MDCN$ is a rectangle. The points $P$ and $E$, $A$ and $B$ are symmetric with axis of symmetry same as for the rectangle $MDCN$ which passes through $O$ and is parallel to $DC$ and $MN$. It follows that $\overline{AP} = \overline{EB}$. From $\overline{CD} = \overline{CE} + \overline{ED}$, $\overline{MD} = \overline{PE} = \overline{AE} - \overline{AP} = \overline{AE} - \overline{EB}$ and because $\angle MDC$ is a right-angled triangle we obtain that

$$d^2 = \overline{MD}^2 + \overline{CD}^2 = (\overline{AE} - \overline{EB})^2 + (\overline{CE} + \overline{ED})^2 = \\ = \overline{AE}^2 + \overline{EB}^2 - 2 \cdot \overline{AE} \cdot \overline{EB} + \overline{CE}^2 + \overline{ED}^2 + 2 \cdot \overline{CE} \cdot \overline{ED} \quad \dots\dots(1).$$

$\angle BAC = \angle BDC$ because they are angles over the same arc $BC$. From the similarity of the triangles $\Delta AEC \sim \Delta BED$ we obtain $\overline{AE} : \overline{CE} = \overline{ED} : \overline{BE}$ or $\overline{AE} \cdot \overline{EB} = \overline{CE} \cdot \overline{ED} \dots(2)$. If we substitute (2) in (1) we get

$$\overline{AE}^2 + \overline{BE}^2 + \overline{CE}^2 + \overline{DE}^2 = d^2.$$