Given, we have an $\square ABCD$ with a side length of $1\text{cm}$. We construct its diagonal $AD$. From $C$, we draw the altitude $CE$ of $\triangle ACD$. Now, we construct the altitude $FE$ of the triangle $\triangle CDE$. Now, we connect $F$ and $A$ and name the intersection point on $CE$ to be $G$. We thus formed an $\triangle ACG$. Then, a circle tangent to all the sides of the $\triangle ACG$ is constructed. What is its area?
EDIT
Well, I could find out some of the lengths of lines and they are as follows:
- $\overline{AD} = \sqrt{2}$
- $\overline{ED} = \frac{1}{\sqrt{2}}$
- $\overline{EF} = \frac{1}{2}$
- $\overline{AF} = \frac{\sqrt{5}}{2}$
Other than this, I have no idea.
We know $AC=1$ which implies $AD=\sqrt{2}$. Next, let we have $AG=a$, $GF=b$, $CG=c\,$ and $\,GE=d$.
Triangle $\triangle ACF$ is right triangle, so we know:
$(a+b)^2 = 1^2 + 0.5^2 \implies \boxed{a+b=\frac{\sqrt{5}}{2}}$.$\qquad (1)$
Triangle $\triangle CFE$ is right triangle, so we know:
$(c+d)^2 = 0.5^2 + 0.5^2 \implies \boxed{c+d=\frac{\sqrt{2}}{2}}$.$\qquad (2)$
Triangle $\triangle AEG$ is right triangle, so we know:
$a^2 = d^2 + (\frac{\sqrt{2}}{2})^2 \implies \boxed{a=\sqrt{\frac{1+2d^2}{2}}}$.$\qquad (3)$
It is obvious that $EF$ bisects right angle $\angle CED$, so we know angle $\alpha = \angle GEF = 45^{\circ}$. Now we can use law of cosines for triangle $\triangle GEF$. We will have:
$b^2=d^2+0.5^2 - 2\cdot d\cdot 0.5\cdot\cos 45^{\circ}=d^2-\frac{\sqrt{2}}{2}d+\frac{1}{4}$, i.e.
$\boxed{b=\sqrt{d^2-\frac{\sqrt{2}}{2}d+\frac{1}{4}}}$.$\qquad (4)$
Adding equations $(3)$ and $(4)$ into equation $(1)$ we can find the value for $d$, i.e. we will find $d \approx 0.2357$. Hence, we can find all other unknowns too. I found $a \approx 0.74536, b \approx 0.37268$ and $c \approx 0.4714$.
Now, since we know all three sides for triangle $\triangle AGC$, we know its perimeter $s=\frac{1+a+c}{2}$ and its area $\Delta=\sqrt{s(s-1)(s-a)(s-c)}$. Hence, we know the radius for its incircle, i.e. $$ \boxed{r=\frac{\Delta}{s} \approx 0.15037}. $$
Finally, the required area is: $$ \boxed{P=r^2\pi \approx 0.07103 \,\, cm^2}. $$