Finding area of sector inside an triangle I have been asked this question from a junior and could not solve the question in a simple way. I am asking help on this platform. 
For a triangle $ABC$, Points $D, E$ on $AB$, where $AD:DE:EB=2:2:1$. Point $H$ on $AC$, where $AH:HC=5:3$. Point $G$ on $BC$, where $BG:GC=1:1$.  Point $F$ is the intersection between $DG$ and $EH$. 
Given area $ADFH$ = 100 $cm^2$, find area $BEFG$. 
The difficulty is I don't see any line parallel, so I can't find any similar triangle. 
What I have derived from triangle area formula is area $AEH$: area $ABC$ = $1:2$ and area $BDG$: area $ABC$ = $3:10$. But to get area $DEF$'s ratio, I spend quite a long time finding the ratio $EF:FH$. 
This should not be such a hard question, so I want to know if there is an easier way of finding the answer.
 A: This is too long for a comment.  Let $\overrightarrow{AB}=a$ and $\overrightarrow{AC}=b$.  Then 
$$0=\overrightarrow{AH}+ \overrightarrow{HF}+\overrightarrow{FD}+\overrightarrow{DA}.$$
Now substitute
$\overrightarrow{AH}=\frac{5}{8}b$,
$\overrightarrow{HF}= x\overrightarrow{HE} =x(-\frac{5}{8}b+\frac{4}{5}a)$, 
$\overrightarrow{FD}=y\overrightarrow{GD}=y\bigl(\frac{1}{2}(a-b)-\frac{3}{5}a\bigr)$ and
$\overrightarrow{DA}=-\frac{2}{5}a$
in that equation to get
$$\left(\frac{4}{5}x-\frac{1}{10}y-\frac{2}{5}\right)a +
\left(\frac{5}{8}-\frac{5}{8}x-\frac{1}{2}y\right)b=0.$$
From here $x=21/37$.
A: The following ratios are assumed:-
(1) area of ⊿AEH : area of ⊿ABC = [⊿AEH] : [⊿ABC] = 1 : 2   {User136705’s effort}
(2) [⊿BGD] : [⊿ABC] = 3 : 10    {User136705’s effort}
(3) EF : FH = 16 : 21       {Michael Hoppe’s effort}
Constructions: Produce BA to X such that XA = EB. Join CX. Join HD. 
Let [⊿CXA] = 1 (square) unit. Then, by same altitude,
(0.1) [⊿CXB] = 6 units
(0.2) [⊿CAB] = 5 units
BG = GC and BD = … = AD implies 
(1.1) DG : CX = 1 : 2 …..{Mid-point theorem}
(1.2) [⊿BGD] = (1/4)[⊿CAB] = … = 3/2 units ….. {ratio of areas of similar objects}
Let $[⊿HDF] = t$ $cm^2$. Then, $[⊿AHD] = 100 – t$ $cm^2$
From (3), $[⊿FDE] = (\frac {16} {21})t$ $cm^2$
∵ $[⊿AHD] = [⊿EHD]$……….{same altitude}
∴ $100 – t = t + (\frac {16} {21})t$
∴ $t = … = 2100/58$ $cm^2$
From (1), $[100 + (\frac {16} {21})(\frac {2100} {58})] cm^2 = (1/2)5$ units
∴ $1$ unit $= (2/5)[100 + \frac {1600} {58}]$ $cm^2$
∴ $[$ quad $EBGF] = [⊿GDB] – [⊿FDE]$
$=(3/2)×(2/5)[100 + \frac {1600}{58}] – (\frac {16}{21})(\frac {2100}{58})$ $cm^2$
$= ... = 48.96551724$ $cm^2$
A: Area of $EBGF=(\det(\overrightarrow{BG},\overrightarrow{BE})+\det(\overrightarrow{EG}, \overrightarrow{EF}))/2$, I'll get for that area $\frac{71}{740}\det(a,b)$.  Similar area of $ADFH=\frac{59}{296}\det(a,b)$.  If that area is supposed to be $100$, area of $EBGF=\frac{2840}{59}\approx 48.14$.
Edit: area of $ADFH=\frac{29}{148}\det(a,b)$, I was off by $1$, so the final result is $1420/29\approx48.97$.
You may achieve easier by noting that the area of $DEF=\frac{2}{37}\det(a,b)$.
