You may see here! http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=361789&hilit=CGMO+2010+p2 Best regards, sunken rock

I have just copied and pasted my post. SKhan wrote: Here is my solution for problem 2. Let the perpendicular bisector of $BE$ and the perpendicular of $DE$ at $D$ meet at the point $P$. $\angle PDE=90^o=\angle PME$ so $PDME$ is a cyclic quadrilateral. Note that $\triangle BPE$ is isosceles so $\angle EPM=\angle MPB$. $D$ and $M$ are the midpoints of $BC$ and $BE$ so $DM\parallel CE$ so $\angle MDB=\angle ECB$ $\angle BPD=\angle MPD-\angle MPB=\angle MED-\angle EPM=\angle BED-\angle EDM$ $\angle BPD=\angle EDB-\angle EDM=\angle MDB=\angle ECB=90^o-\angle DBA=\angle BAD$ Hence $ABDP$ is a cyclic quadrilateral. $P$ and $F$ are on the same side of the diameter $AB$ so $P$ and $F$ coincide. So $ED\perp FD$ as required.

[asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A=(0,120), B=(-40,0), C=(40,0), D=(0,0), E=(-61.4,33.8), F=(41.4,75.2), G=(-18.6,-33.8), H=(-82.8,67.6), M=(-50.7,16.9); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(M); draw(A--B--C--cycle, linewidth(0.5)); draw(circumcircle(A,D,B), linewidth(0.5)); draw(B--E--D, linewidth(0.5)); draw(D--H--E, linewidth(0.4)+dashed); draw(B--G--D, linewidth(0.4)+dashed); draw(D--F--M, linewidth(0.5)); label("$A$", A, (0,1)); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, (0,-1)); label("$E$", E, SW); label("$F'$", F, (1,0)); label("$G$", G, SW); label("$H$", H, (-1,0)); label("$M$", M, SW); draw(C--E, linewidth(0.4)+dashed); draw(H--F, linewidth(0.4)+dashed); [/asy][/asy] Define $F'$ as the point on the perpendicular bisector of $\overline{BE}$ satisfying $\overline{DE}\perp\overline{DF'}$, $G$ as the intersection of $\overline{DF'}$ and $\overline{BE}$, and $H$ as the second intersection of $\overline{BE}$ with $(ABD)$. To begin, we must have that $$\angle BHD=90^\circ-\angle ABD=\angle BCE,$$which used in conjunction with $BD=BE$ implies that $\triangle BCE\cong\triangle BHD$; therefore, $EH=BE$. Furthermore, since $\overline{BD}$ is the median to hypotenuse of $EDG$, it follows that $GB=BE$ as well. Combining our length results with the fact that $EMDF'$ is cyclic yields $$GB\cdot GH=GM\cdot GE=GD\cdot GF',$$which completes our problem by POP.

Let $MF$ meet $AD$ at $X$, let $N$ be the midpoint of $AB$, now notice that $X$ lies on the perpendicular bisector of $EB$ and $BC$ so it is the circumcenter of $ABC$. Since $M,B,D,X$ are concyclic, $$\angle MXB=\angle MDB=\angle ECB=\angle BAD$$Hence $\triangle EXB\sim\triangle BAC$ Moreover, notice that $\angle EBA=\angle EBX-\angle ABX=\angle ABD-\angle ABX=\angle XBD$, and $$\frac{BX}{AB}=\frac{EB}{BC}=\frac{1}{2}$$therefore, $BN=BX$, so B lies on the perpendicular bisector of $NX$ and $ED$ which implies that $NXDE$ is an isoceles trapezoid. We further have $$EX=XB=NB=NF$$and $$\angle NEX=\angle NDX=\angle DAN=\angle DAB=\angle EXM$$So $EN\|MX$, combined with the fact that $E$ and $F$ lies on different side of $NX$, we conclude that $ENFX$ is a parallelogram. Hence $XD=EN=XF$ Now $\angle DXF=\angle MBD$, hence $D$ is the center of spiral similarity sending $EB$ to $XF$, hence also the center of spiral similarity sending $BX$ to $EF$, so $\angle EDF=\angle BDX=90^{\circ}$ This completes the proof.

Let S be where perpendicular at M to BE and perpendicular at D to DE meet. we will show S is F. Clearly SDME is cyclic so ∠MSD = ∠MED = ∠BDE and ∠BSM = ∠MSE = ∠MDE. ∠DSB = ∠DSM - ∠BSM = ∠BDE - ∠MDE = ∠BDM = ∠DAB so DSAB is cyclic and S is F. we're Done.