Problem: In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$
[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label("$A$",A,N); dot(B);  label("$B$", B,SW);dot(C);  label("$C$",C,SE); dot(DD);  label("$D$",DD,NE); dot(EE);  label("$E$",EE,NW); dot(FF);  label("$F$",FF,S); [/asy]
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$

Solution: [asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42;  /* image dimensions */   /* draw figures */ draw(circle((0,0), 5), linewidth(2));  draw((-4,-3)--(4,3), linewidth(2));  draw((-4,-3)--(0,5), linewidth(2));  draw((0,5)--(4,3), linewidth(2));  draw((12,-1)--(-4,-3), linewidth(2));  draw((0,5)--(0,-5), linewidth(2));  draw((-4,-3)--(0,-5), linewidth(2));  draw((4,3)--(0,2.48), linewidth(2));  draw((4,3)--(12,-1), linewidth(2));  draw((-4,-3)--(4,3), linewidth(2));   /* dots and labels */ dot((0,0),dotstyle);  label("E", (0.27,-0.24), NE * labelscalefactor);  dot((-5,0),dotstyle);  dot((-4,-3),dotstyle);  label("B", (-4.45,-3.38), NE * labelscalefactor);  dot((4,3),dotstyle);  label("$D$", (4.15,3.2), NE * labelscalefactor);  dot((0,5),dotstyle);  label("A", (-0.09,5.26), NE * labelscalefactor);  dot((12,-1),dotstyle);  label("C", (12.23,-1.24), NE * labelscalefactor);  dot((0,-5),dotstyle);  label("$G$", (0.19,-4.82), NE * labelscalefactor);  dot((0,2.48),dotstyle);  label("I", (-0.33,2.2), NE * labelscalefactor);  dot((0,0),dotstyle);  label("E", (0.27,-0.24), NE * labelscalefactor);  dot((0,-2.5),dotstyle);  label("F", (0.23,-2.2), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]
Let $A[\Delta XYZ]$ $\text{Area of Triangle XYZ}$
$A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240$
$A[\Delta ABE] = A[\Delta AED] = 60$ (the median divides the area of the triangle into two equal parts)
Construction: Draw a circumcircle around $\Delta ABD$ with $BD$ as is diameter. Extend $AF$ to $G$ such that it meets the circle at $G$ . Draw line $BG$
$A[\Delta ABD] = A[\Delta ABG] = 120$ (Since $\square ABGD$ is cyclic)
But $A[\Delta ABE]$ is common in both with an area of 60. So, $A[\Delta AED] = A[\Delta BEG]$
Therefore $A[\Delta AED] \cong A[\Delta BEG]$ (SAS Congruency Theorem).
In $\Delta AED$ , let $DI$ be the median of $\Delta AED$
which means $A[\Delta AID] = 30 = A[\Delta EID]$
Rotate $\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$ $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\frac{AE}{EF} = 2:1$
$AE$ is a radius and $EF$ is half of it implies $EF$ $\frac{radius}{2}$
which means $A[\Delta BEF] \cong A[\Delta DEI]$
Thus, $A[\Delta BEF] = \boxed{30}$

Answer: B 30