My solution: Let $X=AF \cap BC$ From theorem bisector in $\triangle ABC$: $2BD=DC$ Is easy to see that $(X,D,B,C)=-1$ $\Rightarrow$ $XB=BC$ Also that $AX=AF$. Let $Y=AC \cap FD$ From Theorem Menelao's in $\triangle XAC$: $XD \cdot YC \cdot AF=DC \cdot YA \cdot FX$ $\Rightarrow$ $2DC \cdot YC \cdot AF=DC \cdot YA \cdot 2AF$ $\Rightarrow$ $AY=YC$ Done!!

First let $Y= (AF) \cap (BC) $ and $ X =(FD) \cap (AB)$ Using the well known fact by which $(AD)$ divides $(BC)$ in the ratio $ \frac{AB}{AC}$ we get $DC=2DB$ So we only need $BX=BA$ in order to conclude by Menelaus . But $BX=BA <==> (FA,FX,FB,FC)=-1 <==> (Y,D,B,C)=-1 <==> (AF,AD,AB,AC)=-1$ , wich is true since $AD$ and $AF$ are the internal and external bisectors of $BAC$ . And we're done .

Let say $\angle BAD=\phi$ Then $\angle BAC=\angle ACF=2\phi$ Since $\angle DAF=90^{\circ}$ Then It follows that $\angle CAF=90^{\circ}-\phi$. And It is easy to see that $\triangle CAF$ is isosceles with $2AB=AC=FC$. Let $AB \cap DF =B'$ From Angle Bisector Theorem in triangle $BAC$ we get $BC=k,DC=2k$ Since triagles $BDB'$ and $FDC$ are similar we get $\frac{FC}{BB'}=\frac{DC}{BD}=\frac{2k}{k}=2$ $\longrightarrow$ $2AB=FC=2BB'$ $\longrightarrow$ $AB=BB$' Since triangles $AB'M$ and $FMC$ (Where $M=FD \cap AC$) are similar we get $\boxed{AM=MC}$ as desired .

My solution: From AC=2AB we have $DC=\frac{2}{3}BC$. Let $T$ be intersection of $FA$ and $BC$ and let $P$ be the midpoint of $AC$. Because $\triangle AFC$ is isosceles we have $TA=TF$ and $TB=BC$. We have: $\frac{TD\cdot CP\cdot AF}{DC\cdot PA\cdot TF}=\frac{TD}{2DC}=\frac{2BC-DC}{2DC}=\frac{2BC-\frac{2}{3}BC}{\frac{4}{3}BC}=1$ so by converse Menelaus theorem $F,D,P$ are collinear.

Let $E= FC\cap AD$ and $G$ be the midpoint of $AC$. $\triangle ABG$ is clearly isosceles, therefore $AD$ is the perpendicular bisector of $BG$. $\angle ABG=\angle AGB=\alpha$, and $\angle BAD=\angle DAB=\beta$, where $\alpha+\beta=90^{\circ}$. By angle chasing, $\angle AEF=\beta$, $\angle EFA=\angle CAF=\alpha$. Therefore, $CE=CF=CA$. $\triangle ADB\sim \triangle EDC$, with ratio $1:2$. So, applying Menelaus on $\triangle AEC$: \[\frac{AD}{DE}\cdot \frac{EF}{FC}\cdot \frac{CG}{GA}=1\]Therefore $D, G, F$ collinear.

By the bisector theorem $CD=2DB$ Note that $\angle CAD$+$\angle FAC$$=$$90$ But $\angle FCA$$=$$2\angle CAD$ Therefore $\angle CFA$$=$$90-\angle CAD$ And $\triangle FCA$ is isosceles. The parallel to $FA$ through $C$ intersects $FD$ in $E$ But $\frac{BE}{FC}=\frac{DB}{CD}=\frac{1}{2}$ and $2BE=FC$ and $BE=AB$ And since $AE = FC$ and $FA // CE$ then $FCEA$ is parallelogram and $FE$ and $AC$ intersect at their midpoint since they are diagonal. $\blacksquare$

Let $E$ be the point on the ray $AB$ such that $AE=AC$. Then $\bigtriangleup AEC$ is isosceles and therefore $D$ is it's centroid and $AD \perp EC \Rightarrow AF||EC \Rightarrow AECF$ is a parallelogram. Therefore $D$ lies on $EF$ which bisects $AC\square$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(13); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -17.37511056303312, xmax = 7.5087778813877195, ymin = -2.4450101454350848, ymax = 11.343111648358757; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((-7.031979554700491,5.719760792517097)--(-6.74,-0.6), linewidth(3.6)); draw((-7.031979554700491,5.719760792517097)--(3.0349669446003054,-1.9454268973058424), linewidth(3.6)); draw((3.0349669446003054,-1.9454268973058424)--(-6.74,-0.6), linewidth(3.6)); draw((-16.514966944600314,0.7454268973058432)--(-6.74,-0.6), linewidth(3.6)); draw((xmin, -21.644531922791092*xmin + 63.74501201971122)--(xmax, -21.644531922791092*xmax + 63.74501201971122), linewidth(2) + linetype("4 4")); /* line */ draw((-3.4816776851332314,-1.048475632435281)--(2.451007835199324,10.694094687728354), linewidth(2.8)); draw((-16.514966944600314,0.7454268973058432)--(2.451007835199324,10.694094687728354), linewidth(3.2) + dtsfsf); draw((xmin, 0.5245534651358222*xmin + 2.9354903550154416)--(xmax, 0.5245534651358222*xmax + 2.9354903550154416), linewidth(2.8) + linetype("4 4") + dtsfsf); /* line */ draw((-7.031979554700491,5.719760792517097)--(-3.4816776851332314,-1.048475632435281), linewidth(3.2) + dtsfsf); /* dots and labels */ dot((-7.031979554700491,5.719760792517097),dotstyle); label("$A$", (-7.584728224244594,5.9176081022801155), NE * labelscalefactor); dot((-6.74,-0.6),dotstyle); label("$B$", (-7.034019217687739,-1.3639887621938518), NE * labelscalefactor); dot((3.0349669446003054,-1.9454268973058424),dotstyle); label("$C$", (2.4504136730136468,-2.383820255817657), NE * labelscalefactor); dot((-3.4816776851332314,-1.048475632435281),dotstyle); label("$D$", (-3.586988769239278,-1.8535078791332782), NE * labelscalefactor); dot((2.451007835199324,10.694094687728354),dotstyle); label("$F$", (2.69517323148336,10.833195901546855), NE * labelscalefactor); dot((-1.9985063050500922,1.887166947605627),dotstyle); label("$X$", (-2.2816044574008076,2.3278012447243217), NE * labelscalefactor); dot((-16.514966944600314,0.7454268973058432),dotstyle); label("$G$", (-17.089557744818453,0.8592438939060428), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Define: $DF \cap AC=X$ and $AF \cap BC=G$, from angle bisector theorem, $AC=2AB \implies CD=2BD$, then, $$-1=(G,D;B,C) \Longrightarrow \frac{GB}{BD}=\frac{CG}{CD} \implies GB=BC$$and using the fact, $AB||CF$, $GB=BC$ $\implies$ $GA=AF$, then, $$-1=(G,D;B,C) \overset{X}{=} (G,F;BX \cap FG ,A) \implies BX||FG \implies \frac{CB}{BG}=\frac{CX}{XA}=1$$

Let $X=\overline{AF}\cap \overline{BC}, Y=\overline{AB}\cap\overline{DF}$, and $M$ the midpoint of $AC$. Hence $(X,D;B,C)=-1$. Therefore, $$\frac{XB}{XC}=\frac{DC}{DB}=\frac{AC}{AB}=2$$As $AB\|CF$, so $AY=CF=2AB$, hence $$\frac{AY}{YB}\cdot\frac{BD}{DC}\cdot\frac{CM}{MA}=2\cdot\frac{AB}{AC}\cdot 1=2\cdot\frac{1}{2}\cdot 1=1$$By the converse of Menelaus, $YDMF$ is collinear.