Finding the length of $AE$ In the figure there're $28$ points .The distance between each point is $1$ unit .The segment $AB$ intersect with the segment $CD$ in the point $E$. How to find the length of $AE$

 A: 
$$\bigtriangleup AEF \sim\bigtriangleup BED $$ 
This implies 
 $$  \frac{EB}{AE} = \frac{BD}{AF} $$
 $$  \frac{EB}{AE} = \frac{4}{5} $$  
Add $1$ to both sides 
$$ \frac{AB}{AE} = \frac{9}{5} $$
$$ AE = \frac{5}{9} \times AB $$
$$ AE = \frac{5}{9} \times \sqrt{3^2 + 6^2 } = \frac{5}{9} \times \sqrt{45} $$
$$ AE = \frac{5}{9} \times 3\sqrt{5} $$
$$ \boxed{AE = \displaystyle\frac{5\sqrt{5}}{3}} $$
A: If we treat the bottom left corner as the origin, the two lines have equations:
$$y=-\frac{1}{2}x+3\\y=x-2$$
Solving for the point of intersection gives $(\frac{10}{3},\frac{4}{3})$.  The distance between this point and $(0,3)$ is 
$$\sqrt{\left(\frac{10}{3}\right)^2+\left(\frac{5}{3}\right)^2}=\frac{5\sqrt{5}}{3}$$
A: 
You can calculate $\sin(CAF)$ and $\cos(CAF)$ in $\Delta AFC$. You can also calculate $\sin(FAB)$ and $\cos(FAB)$ in $\Delta AGB$.
Then, 
$$\sin(\EAC)=\sin(CAF-BAG) = \sin(CAF)\cos(BAG)-\cos(CAF)\sin(BAG)$$
As $\angle CDB =45^0$ you can calculate $\sin(ADC)$ in the same way: $ADC+ADG+45^0=180^0$ and $<ADG$ can be calculate in $\Delta ADG$.
Now, use the sin law in $\Delta AED$.
$$\frac{AE}{\sin(ADE)}=\frac{AD}{\sin(AED)}$$
