Let ABCD be a square, E be midpoint of AB. Let F be on BC and G be on CD and EF is parallel to AG. Prove FG is tangent to the incircle of the square. 
I tried using OC and finding the slope so that maybe OC would be perpendicular to FG. I know that EF and AC will have the same slope. I also know that I have to use the distance formula but I am really stuck on how to prove a line is tangent to an incircle.

Let's use $2l$ to be the side of the square, with $O$ at the origin. Then the $F$ point has coordinates $(x,l)$. $G$ has coordinates $(l,y)$. Since $EF||AG$, we can draw the perpendicular from $F$ to the horizontal axis, and we get similar triangles. Then $$\frac{l}{l+x}=\frac{l+y}{2l}$$
This is equivalent of saying that $\tan\angle FEO=\tan\angle GAD$. We can then write $$y=\frac{2l^2}{l+x}-l=\frac{l^2-lx}{l+x}$$
You can now use the formula to calculate the distance from $O$ to $FG$. Simple algebraic manipulation will yield that this distance is $l$. Draw the perpendicular from $O$ to $FG$. Since this distance is $l$ it means that $FG$ is tangent to the circle of radius $l$ centered in $O$, which is the incircle of the square.
