Square, Triangle, And Semicircle Living In Harmony

Let M be the midpoint of AD. H=CM $\cap$ FD . Sides of the square 2k. So square area = $\color{#3D99F6}{4k^2=12.25}$
CF and CD are tangents to semicircle AFED with center M.
$\implies ~\Delta ~ CFD ~is ~ isosceles.~ \Delta s~ CMD, ~ CHD ~ are ~\color{#3D99F6}{ rt.~\angle ed.}\\ \color{#3D99F6}{Semidiagonal ~ ED=\sqrt2k, ~~MD=k, CD=2k ~~CM=\sqrt5k.}\\ \angle ~EDC~=45^o, \text{ angle between a diagonal and the side of a square.}\\ \implies~\angle DCM= \alpha=Sin^{-1}\dfrac 1{\sqrt5},~~~ ~ \angle HDC=\angle HDE+45^o= Cos^{-1}\dfrac 1{\sqrt5}\\\color{#3D99F6}{ \implies ~\beta= Cos^{-1}\dfrac 1{\sqrt5} - 45^o, }\\ Sin\beta=Sin(Cos^{-1}\dfrac 1{\sqrt5})*Cos45-Cos(Cos^{-1}\dfrac 1{\sqrt5})*Sin45=\color{#3D99F6}{\dfrac 1 {\sqrt2*\sqrt5}}\\ FD=2*CDSin\alpha=2*2*\dfrac 1{\sqrt5}.\\ Area~ \Delta ~ EFD =\frac 1 2 *FD*k*EDSin\beta*k \\ \frac 1 2*(4*\dfrac 1{\sqrt5})*(\sqrt2)*(\dfrac 1 {\sqrt2*\sqrt5}) *\dfrac{12.25} 4 \\ Area~ \Delta ~ EFD =1.225.\\$

Let $M$ denote the midpoint of $AD$ , observe that $DEF\sim CMA$ (Use the fact that $A,F,E,D$ are concyclic.) Thus by ratio of areas of similar shapes we have: $[DEF]=[CAM](\frac {DE}{CM})^2=\boxed {1.225}$ .

The equation of the circle is $\left(x-\frac{3.5}{2}\right)^2+y^2=\frac{12.25}{4}$ .

The equation of the diagonal is $y=-\left(x-3.5\right)$

The equation of the tangent line can be found by assuming it to be $y-3.5=m(x-3.5)$ and then solving for $m$ by using the fact that there is only one solution to the system of equations involving the tangent line and the circle.Hence, $m=0.75$

Then we can use all these equations and get the co-ordinates of the vertices of the triangle and find it's area.

AF is perpendicular to FD and since triangles MFC & MDC are congruent with MF=MD,CF=CD & MC common, so MC is also perpendicular to FD, In other words, AF//MC and of course A, E, C are collinear. This should make it easy to see the similarity.