The diagonals of the square $ABCD$ intersect at $P$ and the midpoint of the side $AB$ is $E$. Segment $ED$ intersects the diagonal $AC$ at point $F$ and segment $EC$ intersects the diagonal $BD$ at $G$. Inside the quadrilateral $EFPG$, draw a circle of radius $r$ tangent to all the sides of this quadrilateral. Prove that $r = | EF | - | FP |$.
WLOG, let the side length of the square be $1$. Let $A=(0,6), B=(6,6), C=(6,0),$ and $D=(0,0)$. We find $P=(3,3)$. By coordbashing, $F=(2,4)$ and $G=(4,4)$. Our last point is $E=(3,6)$ We use pythag. to get $EF=EG=\sqrt5$. Next, $PF=PG=\sqrt2$. We are now trying to prove that the radius $r$ of the incircle $EFPG$ is $\sqrt5-\sqrt2$. We set point $I$ as the incenter of $EFPG$. The radii from $I$ to $PF$ and $FE$ have feet at $X$ and $Y$, repsectively. $XF=\sqrt2-r$ (there is a square formed if you draw a diagram) and $XF=FY$ because $IXFY$ is a kite. Thus, $FE=\sqrt5-\sqrt2+r$. We find $IP=r\sqrt2$, so $PE=3-r\sqrt2$. All that's left is to prove is that if $r=\sqrt5-\sqrt2$, then $IYE$ is a right triangle with legs $r$ and $\sqrt5-\sqrt2+r$ and hypotenuse $3-r\sqrt2$.