Prove that $DE\perp EF$. 
Question: Point $D$ lies inside $\Delta ABC$ such that $\angle DAC=\angle DCA=30^\circ$ and $\angle DBA = 60^\circ$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE\perp EF$.

My approach: Let $\angle CDF=\alpha$ and $\angle EDC=\beta$. 
Now in $\Delta FDA$, we have $$\frac{FD}{\sin 30^\circ}=\frac{FA}{\sin(120^\circ-\alpha)}\\\implies FD=\frac{1}{2}.\frac{FA}{\sin(120^\circ-\alpha)}.$$
Again in $\Delta FDC$, we have $$\frac{FD}{\sin 30^\circ}=\frac{FC}{\sin \alpha}\\\implies FD=\frac{1}{2}.\frac{FC}{\sin \alpha}.$$
Thus, $$\frac{1}{2}.\frac{FA}{\sin(120^\circ-\alpha)}=\frac{1}{2}.\frac{FC}{\sin \alpha}\\\implies \frac{\sin \alpha}{\sin(120^\circ-\alpha)}=\frac{FC}{FA}=\frac{1}{2}\\\implies \tan \alpha=\frac{1}{\sqrt 3}\implies \alpha=30^\circ.$$
Thus $\angle ADF=90^\circ$. Now let $CD$ extended meet $AB$ at $J$. Thus $\angle ADJ=60^\circ.$ Now observe that if we can prove that points $A,D$ and $E$ are collinear, then we can conclude that $\angle EDC=\beta=60^\circ$. Hence we will be done. 
I tried to use Menalaus Theorem to prove the same, but it was of no use.
Also I tried to use coordinate bash. Consider $\Delta CDA$. Observe that $\Delta CDA$ is isosceles with $CD=BA$. Let $DO$ be the angular bisector of $\angle ADC$. Thus $DO$ is also the perpendicular bisector of $AC$. Now let $O$ be the origin and let $AC$ be the x-axis. Thus clearly $DO$ represents the y-axis. Now let $DA=s$. Thus clearly $A=\left(-\frac{\sqrt 3}{2}s,0\right),C=\left(\frac{\sqrt 3}{2}s,0\right)$ and $D=\left(0,\frac{s}{2}\right)$. Now let $B=(a,b)$, thus $$E=\left(\frac{a}{2}+\frac{\sqrt{3}}{4}s, \frac{b}{2}\right).$$ 
Now slope of $AB=m_1=\frac{2b}{2a+\sqrt 3s}$ and slope of $DB=m_2=\frac{2b-s}{2a}$. Now since the angle between $AB$ and $DB=60^\circ,$ thus we have $$\sqrt 3=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.$$ After this I haven't found anything significant. 
So, how to proceed after this?
 A: Arbitrarily set $A=(0,0)$, $C=(1,0)$, $D=\left(\frac12,\frac1{2\sqrt3}\right)$ and $F=\left(\frac23,0\right)$. Now define new points $O=\left(\frac16,\frac1{2\sqrt3}\right)$, $G=\left(\frac13,0\right)$ and $H=\left(0,\frac1{\sqrt3}\right)$, then let $\Gamma$ be the circle with centre $O$ passing through $A$.
Since $\angle DOA=120^\circ$, $B$ lies on $\Gamma$. We also have the equalities
$$GF=FC,BE=EC,HD=DC$$
so we can show $\triangle HBG$ is similar to $\triangle DEF$. Now $GH$ happens to be a diameter of $\Gamma$, so $\angle HBG=90^\circ$ and hence $\angle DEF=90^\circ$ too, or $DE\perp EF$.
A: Let $AD=DC=p$ and $\measuredangle BAD=\alpha$.
Thus, $$AC=p\sqrt3,$$
$\frac{BD}{\sin\alpha}=\frac{p}{\sin60^{\circ}},$ which gives $$BD=\frac{2p\sin\alpha}{\sqrt3},$$ $\frac{AB}{\sin(60^{\circ}+\alpha)}=\frac{p}{\sin60^{\circ}},$ which gives $$AB=\frac{2p\sin(60^{\circ}+\alpha)}{\sqrt3}$$ and 
$$\vec{DE}\cdot\vec{FE}=\frac{1}{2}\left(\vec{DB}+\vec{DC}\right)\left(\frac{1}{3}\vec{AC}+\frac{1}{2}\left(-\vec{AC}+\vec{AB}\right)\right)=$$
$$=\frac{1}{12}(\vec{DB}+\vec{DC})(3\vec{AB}-\vec{AC})=\frac{1}{12}\left(3\vec{DB}\cdot\vec{AB}-\vec{DB}\cdot\vec{AC}+3\vec{DC}\cdot\vec{AB}-\vec{DC}\cdot\vec{AC}\right)=$$
$$=\frac{1}{12}\left(3\cdot\frac{2p\sin\alpha}{\sqrt3}\cdot\frac{2p\sin(60^{\circ}+\alpha)}{\sqrt3}\cdot\cos60^{\circ}-\frac{2p\sin\alpha}{\sqrt3}\cdot p\sqrt3\cdot\cos(90^{\circ}+\alpha)\right)+$$
$$+\frac{1}{12}\left(3\cdot p\cdot\frac{2p\sin(60^{\circ}+\alpha)}{\sqrt3}\cdot\cos(60^{\circ}+\alpha)-p\cdot p\sqrt3\cdot\cos30^{\circ}\right)=$$
$$=\frac{p^2}{12}\left(2\sin\alpha\sin(60^{\circ}+\alpha)+2\sin^2\alpha+2\sqrt3\sin(60^{\circ}+\alpha)(\cos(60^{\circ}+\alpha)-\frac{3}{2}\right)=$$
$$=\frac{p^2}{12}\left(\cos60^{\circ}-\cos(60^{\circ}+2\alpha)+1-\cos2\alpha+\sqrt3\sin(120^{\circ}+2\alpha)-\frac{3}{2}\right)=$$
$$=\frac{p^2}{12}\left(-2\cos30^{\circ}\cos(30^{\circ}+2\alpha)+\sqrt3\sin(120^{\circ}+2\alpha)\right)=0$$ and we are done! 
A: 
Let $M$ and $G$ be the midpoints of $AC$ and $AF$, respectively.  If $h$ is the homothety about $C$ with dilatation ratio $2$, then we see that
$$h(F)=G\,,\,\,h(M)=A\,,\text{ and }h(E)=B\,.$$
Let $\omega$ and $\Omega$ denote the circumcircles of the triangles $DFM$ and $DGA$, respectively.  
First, if $D'$ is the reflection of $D$ about the line $AC$, then the triangle $ADD'$ is an equilateral triangle.  Since $AM$ is a median of the triangle $ADD'$ with $AG:GM=2:1$, we see that $G$ is the centroid of the triangle $ADD'$.  Since the centroid of an equilateral triangle is also its circumcenter, we get that $GD=GA$.  Similarly, $FD=FC$.  As $CF=FG=GA$, we get
$$FD=FG=GD\,;$$
therefore, $DFG$ is an equilateral triangle.  Thus, $\angle DGA=120^\circ$.  As $\angle DBA=60^\circ$, we conclude that $DGAB$ is a cyclic quadrilateral.  Thus, $B\in \Omega$.
Extend $CD$ to meet $\Omega$ again at $H$.  Then, $$\angle ADH=180^\circ-\angle ADC=180^\circ-120^\circ=60^\circ\,.$$
Furthermore, $$\angle DHA=\angle DBA=60^\circ\,.$$
Thus, $DHA$ is an equilateral triangle, whence $\angle DAH=60^\circ$. This means
$$\angle GAH=\angle GAD+\angle DAH=30^\circ+60^\circ=90^\circ=\angle FMD\,.$$
This shows that $h(D)=H$, and so
$$h(\omega)=\Omega\,.$$
Now, since $B\in \Omega$, $h(E)=B$, and $h(\omega)=\Omega$, we conclude that $E\in\omega$.  Thus, $DEFM$ is a cyclic quadrilateral.  Because $\angle DMF=90^\circ$, we deduce that $\angle DEF=90^\circ$ as well.  Ergo, $DE\perp EF$.
