<s>[INST] 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.
 [/INST] Line $EF$ has the parametric form $\pmatrix {-1 \\ 0}+\mu\pmatrix {1+x \\ 1}$.
Line $AG$ has the parametric form $\pmatrix {-1 \\ -1}+\lambda\pmatrix {1+x \\ 1}=\pmatrix {1 \\ y}$.
$\therefore\lambda=\dfrac{2}{1+x}$
$\therefore y=-1+\dfrac{2}{1+x}=\dfrac{1-x}{1+x}$
Line $FG$ has the parametric form $\pmatrix {x \\ 1}+\alpha\pmatrix {1-x \\ -1+\dfrac{1-x}{1+x}}=\pmatrix {x \\ 1}+\alpha\pmatrix {1-x \\\dfrac{-2x}{1+x}}$.
The length of line $FG$ is $\sqrt{(1-x)^2 + \left(\dfrac{2x}{1+x}\right)^2} = \dfrac{1+x^2}{1+x}$.
Expand $\triangle FCG$ so that the hypotenuse is $1$ (i.e. multiply all sides by $\dfrac{1+x}{1+x^2}$).
Place the corner $C$ at the origin of the circle, with the $y$ side along the $x$-axis. This touches the circle, and also $\triangle FCG$ at $90^\circ$ (because the rotation is $90^\circ$), and when
$x+a(1-x)=\dfrac{2x}{1+x^2}$
$a(1-x)=\dfrac{x-x^3}{1+x^2}$
$a=\dfrac{x(1+x)}{1+x^2}$
we have that
$1+\dfrac{x(1+x)}{1+x^2}\left(\dfrac{-2x}{1+x}\right)=\dfrac{1-x^2}{1+x^2}$.
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