The triangle $ABC$ is equilateral, because $AB=BC$ and $\angle{ABC}=60$, and the triangle $ACD$ is equilateral too, because $AD=DC=AC$. Now, $CD\|AB$ and $BC\|AD$, so $ADE$ and $DCE$ are similar, so $\frac{AE}{AD}=\frac{DC}{CF}$, but $ADC$ is equilateral, so $\frac{AE}{AC}=\frac{AC}{CF}$. Also, we note that $\angle{EAC}=\angle{ACF}=120$, so $EAC$ and $ACF$ are similar, so $\angle{AEC}=\angle{CAF}$, so $CA$ is tangent to the circuncircle of $AME$. Hence $CA^{2}=CM\times{CE}$.

Let $AB=1$ and consider the inversion $I$ of pole $C$ and power 1. Then $I(E)=E'$ is the intersection of line $CE$ and the circumcircle of $\triangle{ABC}$. Since $\triangle{EAD}\sim{\triangle{DCF}}$ it follows that $AE=\frac{1}{CF}=CF'$, where $F'$ is the inverse of point $F$. Thus $\triangle{BEF'}$ is equilateral and consequently points $A,E',F$ are collinear,i.e. $E'=M$. Considering this we get $AC^2=1=CM\cdot{CE}$.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO1993Problem1 Vo Duc Dien

This is actually very hard for an APMO 1. Here is a very long solution which uses only the most ultra basic techniques: We draw the line segment connecting points $A$ and $C$. Since quadrilateral $ABCD$ is a rhombus with $\angle ABC=60^\circ$, $\triangle ABC$ and $\triangle ADC$ are both equilateral. Hence, $AB=BC=AC=AD=DC$. Let the intersection of $\overline{AD}$ and $\overline{EC}$ be $H$ and let the intersection of $\overline{AF}$ and $\overline{CD}$ be $I$. Claim: $HD=IC$. Proof. Since $AD \parallel CF$ and $AE \parallel CD$ and $E,D,$ and $F$ are colinear, $\triangle EAD \sim \triangle DCF$ by AA Similarity. Therefore, $\frac{AD}{CF} = \frac{ED}{DF} \Rightarrow CF \cdot ED = AD \cdot DF$. Similarly, $\triangle EHD \sim \triangle ECF$ by AA Similarity. Therefore, $\frac {HD}{CF} = \frac{ED}{EF}\Rightarrow HD = \frac{CF \cdot ED}{EF}$. We substitute $CF \cdot ED = AD \cdot DF$ into $HD = \frac{CF \cdot ED}{EF}$ to get $HD = \frac{AD \cdot DF}{EF}~~~(1)$. Since $CD \parallel BE$, $\triangle FCD \sim \triangle FBE$ by AA Similarity. Therefore, $\frac{FC}{FB} =\frac{FD}{FE}$. Similarly, $\triangle FIC \sim \triangle FAB$ by AA Similarity. Therefore, $\frac{IC}{AB} = \frac{FC}{FB} \Rightarrow IC=\frac{FC\cdot AB}{FB}~~~(2)$. Equations $(1)$ and $(2)$ are equal since $AD=AB$ from the equilateral triangles and since $\frac{FC}{FB}=\frac{FD}{FE}$ from $\triangle FIC \sim \triangle FAB$. Hence, $HD = \frac{AD \cdot DF}{EF}=\frac{FC\cdot AB}{FB}=IC$. $\square$ Because $\triangle ACD$ is equilateral, we have $AC=CD$ and $\angle ACD = \angle CDA$. But we also found that $HD=IC$. Thus, $\triangle CIA \cong \triangle DHC$ by SAS Congruence. Let $\angle ECD = x$. It follows that $\angle ECA=180^\circ-\angle BCA - \angle ECD - \angle DCF=180^\circ-60^\circ-x-60^\circ=60^\circ-x$. But it also follows that $\angle CAI=\angle CAM=x$ because $\triangle CIA \cong \triangle DHC$. Therefore, $\angle AMC = 180^\circ-\angle CAM - \angle ACM =180^\circ-x-(60^\circ-x)=120^\circ$. So $\triangle AMC$ has angle measures $\angle A=x, \angle M=120^\circ,$ and $\angle C = 60^\circ-x$. Now we look at $\triangle EAC$. We already know that $\angle ACE=60^\circ-x$. Because $\triangle EHA \sim \triangle CHD$, we have $\angle HEA=x$ and $\angle HAE=60^\circ$. It follows that $\angle CAE=120^\circ$. Hence, $\triangle EAC$ has angle measures $\angle E=x, \angle A=120^\circ,$ and $\angle C=60^\circ-x$. By AAA Similarity, we have $\triangle EAC \sim \triangle AMC$ so $\frac{CA}{CM}=\frac{CE}{CA}\Rightarrow CA^2=CM\cdot CE$ as desired. $\blacksquare$

Sailor wrote: Let $AB=1$ and consider the inversion $I$ of pole $C$ and power 1. Then $I(E)=E'$ is the intersection of line $CE$ and the circumcircle of $\triangle{ABC}$. Since $\triangle{EAD}\sim{\triangle{DCF}}$ it follows that $AE=\frac{1}{CF}=CF'$, where $F'$ is the inverse of point $F$. Thus $\triangle{BEF'}$ is equilateral and consequently points $A,E',F$ are collinear,i.e. $E'=M$. Considering this we get $AC^2=1=CM\cdot{CE}$. 15 years later...................

[asy][asy] import olympiad; markscalefactor=0.01; size(8cm); defaultpen(fontsize(10)); pair D = (0,0); pair C = (1,0); pair B = C+dir(60); pair A = D+B-C; pair E = (-0.8,0.866025403784439); pair F = extension(E,D,B,C); pair M = extension(E,C,A,F); filldraw(A--B--C--D--cycle,opacity(1)+lightcyan, blue); draw(A--E--C--cycle,paleblue+linewidth(1.0pt)); draw(C--A--F--cycle,lightgreen+linewidth(1.0pt)); draw(E--F); label(A,"$A$",NW); label(B,"$B$",E); label(C,"$C$",dir(0)); label(D,"$D$",dir(-150)); label(E,"$E$",dir(-180)); label(F,"$F$",dir(0)); label(M,"$M$",dir(40)); draw(circumcircle(A,M,E),dotted); filldraw(anglemark(A,D,E),purple,heavymagenta+linewidth(0.5)); filldraw(anglemark(C,F,D),purple,heavymagenta+linewidth(0.5)); filldraw(anglemark(D,E,A),orange,red+linewidth(0.5)); filldraw(anglemark(F,D,C),orange,red+linewidth(0.5)); filldraw(anglemark(E,A,C),pink,magenta+linewidth(0.5)); filldraw(anglemark(A,C,F),pink,magenta+linewidth(0.5)); dot(A^^B^^C^^M); [/asy][/asy] It is clear that because $ABCD$ is a rhombus, $AB=BC$ combined with $\angle ABC=60^\circ$ yields that $\triangle ABC$ is equilateral, and analogously that $\triangle ADC$ is equilateral. So, $\angle EAC=180^\circ-\angle CAB=120^\circ,$ and by symmetry $\angle ACF = \angle EAC = 120^\circ.$ By the converse of Power of a Point, proving $AC^2 = CE \cdot CM$ is equivalent to proving that the circumcircle of $\triangle AME$ is tangent to $AC,$ which is equivalent to proving that $\angle AEC=\angle CAF.$ Note that $\angle EDA=\angle DFC$ since $\overline{AD} \parallel \overline{BF}$, and $\angle AED= \angle CDF$ since $\overline{EB} \parallel \overline{DC}.$ So $\triangle DAE \sim \triangle FCD$ by AA similarity, which implies that$$\frac{DA}{FC}=\frac{AE}{CD}.$$Let the side length of the rhombus be $s.$ Since $DA=CD=AC=s,$ the previous equations becomes$$\frac{s}{FC}=\frac{AE}{s} \implies \frac{AC}{FC}=\frac{AE}{AC}.$$However, $\angle EAC=\angle ACF,$ so $\triangle AEC \sim \triangle CAF$ by SAS similarity, which yields the desired $\angle AEC=\angle CAF.$

Claim: Triangles $EAC$ and $AMC$ are similar. Proof: Note that by proving this claim, we have completed the problem, since \[\frac{CM}{AC} = \frac{AC}{CE} \implies CM \cdot CE = AC^2.\]By construction, we know rhombus $ABCD$ is formed from joining equilateral triangles $\triangle ABC$ and $\triangle ACD.$ Furthermore $AD \parallel CF$ and $AE \parallel CD$ since $CF$ is an extension of side $BC$ and $AE$ is an extension of side $BA.$ This implies $\angle AED = \angle CDF$ and $\angle ADE = \angle CFD,$ so by AA similarity, $\triangle EAD \sim \triangle DCF.$ However, recall that $AC=AD=CD$ due to properties of an equialteral triangle, so $\tfrac{EA}{AC}=\tfrac{AC}{CF}.$ Furthermore, as $\angle EAC = \angle ACF = 120^\circ,$ it follows $\triangle CAE ~ \triangle FCA.$ From this similarity, we deduce $\angle AEC = \angle CAF.$ Once again we see another pair of similar triangles. Noting $\angle AEC = \angle CAF$ and triangles $EAC$ and $AMC$ both share angle $\angle ACM,$ we obtain $\triangle EAC ~ \triangle AMC,$ as desired.