28th Hellenic Mathematical Olympiad

Direct. Let $MZ\perp BC$. Then $MZ\parallel AD$, since $AD\perp BC$. Therefore we have: $Z\hat{M}C=E\hat{A}C$. From the quadrilateral $EANC$ we have: $E\hat{A}C=E\hat{N}C$ and hence we get $$Z\hat{M}C=E\hat{A}C=E\hat{N}C=Z\hat{N}C.$$ Therefore the quadrilateral $MNCZ$ is cyclic and then it follows $$A\hat{C}Z=M\hat{C}Z=M\hat{N}Z=B\hat{N}E.$$ Figure 3 Also we have $B\hat{N}E=B\hat{A}E$, (from the cyclic quadrilateral $ABEN$). Therefore we have $$A\hat{C}Z=B\hat{A}E=B\hat{A}D.(1)$$ From the right angled triangle $ABD$ we have $$B\hat{A}D=90^{\circ }-A\hat{B}D=90^{\circ }-A\hat{B}C,(2)$$ and then $$A\hat{C}Z=90^{\circ }-A\hat{B}C.(3)$$ Since $O$ is the circumcentre of the acute angled triangle $ABC$ we get $$A\hat{C}O=90^{\circ }-A\hat{B}C.(4)$$ From (3) and (4) we obtain $A\hat{C}Z=90^{\circ }-A\hat{B}C=A\hat{C}O$ and since the points $Z$ and $O$ lie on the same half plane with respect to $AC$, it follows that the points $C$, $O$ and $Z$ are collinear. Therefore we distinguish the cases: $O\equiv Z$, and then we have our result. $O\not\equiv Z$, and then the straight line $OZ$ is the perpendicular bisector of the side $AB$. Since $C$ belongs to $OZ$, we conclude that $CA=CB$.

Converse If $CA=CB$ or $O\equiv Z$, then the points $C$, $O$ and $Z$ are collinear, because they belong to the perpendicular bisector of the side $AB$. Hence $$A\hat{C}Z=A\hat{C}O=90^{\circ }-A\hat{B}C=B\hat{A}D=B\hat{A}E=B\hat{N}E=M\hat{N}Z\Rightarrow M\hat{C}Z=M\hat{N}Z,$$ and so the quadrilateral $MNCZ$ is cyclic. It gives that $$Z\hat{M}C=Z\hat{N}C=E\hat{N}C=E\hat{A}C.$$ Therefore $ZM\parallel AE$ and since $AE\perp BC$, it follows that $ZM\perp BC$.