The ratio of two areas of triangles On the picture below is a square. The question is to find $s/r$.
I tried to draw a line from the middle of one side of the square to the middle of the opposite one, and it appears that the area of $s$ is $1/4$ the area of our square. So I stopped to calculate the ratio of triangle's area $r$ to the area of whole square.
Any hints?

 A: 
Denote areas [.] and [ABCD] = $I$. Per the Ceva's theorem on  triangle ABD,  $\frac{AF}{FD} = \frac{BI}{ID}\frac{AE}{EB}=\frac{BJ}{AD}\frac{AE}{EB}=\frac12\frac11=\frac12$. Evaluate the ratios below
$$\frac{HC}{FC}=\frac{[DCE]}{[DCEF]}=\frac{[DCE]}{I-[BCE]-[FAE]}
=\frac{\frac12I}{(1-\frac14-\frac14\frac13)I}=\frac34
$$
$$\frac{DG}{ED}=\frac{[ADJ]}{[AEJD]}=\frac{[ADJ]}{I-[CDJ]-[BEJ]}
=\frac{\frac12I}{(1-\frac14-\frac14\frac12)I}=\frac45
$$
$$\frac{DH}{ED}=\frac{[FDC]}{[FDCE]}=\frac{[FDC]}{I-[BCE]-[FAE]}
=\frac{\frac13I}{(1-\frac14-\frac14\frac13)I}=\frac12
$$
Then,
$\frac{HG}{DE}=\frac{DG}{DE}- \frac{DH}{DE} =\frac45-\frac12= \frac3{10}
$ and the area ratio is
$$\frac{[S]}{[R]} =\frac{\frac{HC}{FC}[CDF]}{\frac{HG}{DE}[DEF]}
=\frac{\frac34(\frac13I)}{\frac3{10}(\frac14\frac23I)}=5
$$
A: I don't know if this is the easiest way.  It certainly isn't particularly elegant.
I made the assumption that the side length of the square is 1.
I then found the coordinates of the points on the corners of R.
We can find the equations of two lines that make the sides of R
$y - \frac 12 x\\
y + 2x = 1$
And solve for the point of intersection.
$(x,y) = (\frac 25, \frac 15)$
Now we have a line though this point and the point $(1,0)$
$3y + x = 1$
Which hits the side at $(0, \frac 13)$
Then there is a side with a line through this line at the point (1,1)
$3y - 2x = 1$ which intersects the side whose equation we already have $y+2x = 1$
$4y = 2$
$(x,y) = (\frac 14, \frac 12)$
The points on the corners of $R: (\frac 25,\frac 15),(0,\frac 13),(\frac 14,\frac 12)$
Using the shoelaces algorithm we get the Area of $R = \frac {1}{20}.$
S has base 1, height $\frac 12$
Area of $S = \frac 14$
$\frac {\text {Area of S}}{\text{Area of R}} = \frac {5}{1}$
A: This is the simplest solution I could come up with. Let's first observe that if we find the following ratios,
$$\frac{AH}{HI} \;and \;\frac{BH}{HG}$$
we will be done (the reason is the well-known area distribution according to the bases).

Let one side of the square be $24k$ units (24 because it is divisible by many numbers). So, $GF$ and $FE$ are $12k$.
First, from many different methods involving similar triangles it can be noted that (for example: connect $AE$ and $DG$ and name their intersection $O$, find $AO/OE$ and use Ceva's theorem in $\bigtriangleup ACE$ ):
$$\frac{AB}{BC}=\frac{2}{1}$$
Therefore, $BC=8k$ and $AB=16k$.
From point $B$, let's draw a line that is parallel to the upper (and lower) sides of the square, and name this point $B'$. From the following similarities, the ratio of $BH$ to $HG$ can be deduced:
$$\bigtriangleup ABB' \sim \bigtriangleup ACD \;\; and \; \bigtriangleup AHG \sim \bigtriangleup B'HB \implies \frac{BH}{HG}=\frac{1}{3}$$
Let's now use Menelaus' Theorem in the following way:
$$\frac{ED}{ED+DC}\cdot\frac{CB}{BA}\cdot\frac{AI}{ID}=1 \implies \frac{12}{12+12}\cdot\frac{8}{16}\cdot\frac{AI}{ID}=1\implies \frac{AI}{ID}=\frac{4}{1}$$
From point $H$, let's draw a line that is parallel to the upper (and lower) sides of the square, and name this point $H'$. Observe:
$$\bigtriangleup BAG \sim \bigtriangleup BH'H \;\; and \; \bigtriangleup AH'H \sim \bigtriangleup ACD \implies \frac{AH}{HD}=\frac{1}{2}$$
This shows us that point $H$ is actually on the midsegment of this square.
Combining the ratios obtained for $AI/ID$ and $AH/HD$:
$$\frac{AH}{HI}=\frac{3}{5}$$
Finally, $$\frac{5}{3}\cdot\frac{3}{1}=5 \; \; \blacksquare $$
These types of area ratio problems are fun, see CTMC (q. 25).
