As shown in the figure, $ABCD$ is a square, $E$ is the mid-point of $AB$. The circle with its center at $H$ is tangent with $AD$, $AE$ and $DE$. The circle with its center at $F$ is tangent with $BC$, $BE$ and $DE$. The circle with its center at $G$ is tangent with $BC$, $CD$ and $DE$. Prove that $\triangle{FGH}$ is a right triangle. Click here for the proof.
Proof: Extend $DE$ and $CB$ so that they intersect at $K$. Draw diagonals of $ABCD$ so that they intersect at $O$. Let $M$ and $N$ are tangent points of circle $F$ on $DE$ and $AB$ respectively, we have $FM\perp DE$ and $FN\perp AB$. Additionally, without loss of generality, we assume that $ABCD$ is a unit square. It is obvious that $F$ is on $BD$, $G$ and $H$ are on $AC$.
Let the radius of circle $F$, $G$, $H$ as $f$, $g$, $h$ respectively. We have ${DF}^2={FM}^2+{DM}^2$
We have $$((1-f)\sqrt{2})^2=(\dfrac{\sqrt{5}-1}{2}+f)^2+f^2$$
Solve the above equation, we have $$f=\dfrac{\sqrt{5}-1}{4}$$
Since the perimeter of $\triangle{DAE}$ is $\dfrac{3+\sqrt{5}}{2}$, and its area is $\dfrac{1}{4}$. Therefore, $$h=\dfrac{2\cdot\dfrac{1}{4}}{\dfrac{3+\sqrt{5}}{2}}=\dfrac{3-\sqrt{5}}{4}$$
Since $\triangle{DAE}\sim \triangle{KCD}$, and $CD=2{AE}$, therefore $$g=2h=\dfrac{3-\sqrt{5}}{2}$$
It is easy to show that ${FO}^2={GO}\cdot{HO}$, therefore, $\triangle{FGH}$ is a right triangle.
