A small red segment and an angle

It’s not going to be short, so here is an outline of the steps that follow:

Step 1 : $A$ , $B$ , $M$ and $N$ are homocyclic points.
Step 2 : $N$ is the center of the circumscribed circle of $ABCD$ .
Step 3 : $\angle CBD = 18^\circ$ (the orange one), hence $MN$ and $CD$ can be considered as sides of inscribed regular decagons.
Step 4 : Relate the side of a regular decagon with its circumradius.
Step 5 : Calculate the lengths of the radii of the two circles.
Step 6 : Calculate the area of $ABCD$ as the sum of the areas of $\vartriangle ABD$ and $\vartriangle BCD$ .

Step 1 :
$\angle BMA = 90^\circ = \angle BNA$ , hence $A$ , $B$ , $M$ and $N$ are homocyclic points (circle $c_1$ ) and $AB$ is a diameter of this circle.

Step 2 :
$BD$ is a diameter of the circumscribed circle of $ABCD$ (circle $c_2$ ), hence $\angle BAD = \angle BCD = 90^\circ$ .
${\theta _1}^\circ = {\theta _2}^\circ$ (inscribed angles subtended by the same arc $\overset\frown{AD}$ ).
${\theta _2}^\circ = {\theta _3}^\circ$ (similar reason).
Hence, ${\theta _1}^\circ = {\theta _3}^\circ$ .

Consequently, $\angle MCE = 90^\circ - {\theta _1}^\circ = 90^\circ - {\theta _3}^\circ = \angle CME$ . Since $\angle MCE$ and $\angle CME$ are the acute angles of the right-angled $\vartriangle EMC$ , they both equal to $45^\circ$ . Therefore, ${\theta _2}^\circ = {\theta _1}^\circ = 90^\circ - {\omega _1}^\circ = 90^\circ - 45^\circ = 45^\circ$ and so the right-angled $\vartriangle ABD$ is an isosceles triangle. Since $AN$ is a height of $\vartriangle ABD$ , it is the perpendicular bisector of $BD$ and $N$ is the midpoint of diameter $BD$ , thus it is the center of the circle $c_2$ .

Step 3 :
On $\vartriangle FCD$ : $\angle D + \angle C + \angle F = 180^\circ \Rightarrow \angle D + 45^\circ + 63^\circ = 180^\circ \Rightarrow \angle D = 72^\circ$
On $\vartriangle BCD$ : $\angle B + \angle C + \angle D = 180^\circ \Rightarrow \angle B + 90^\circ + 72^\circ = 180^\circ \Rightarrow \angle B = 18^\circ$

Since $\angle CBD = 18^\circ$ , the chords $MN$ and $CD$ can be considered as sides of inscribed regular decagons in circles $c_1$ an $c_2$ respectively.

Step 4 :

THEOREM
The relation between the side $s$ of a regular decagon and its circumradius $R$ is $R = s \cdot \varphi$ , where $\varphi = \frac{{\sqrt 5 + 1}}{2}$ is the golden ratio.

PROOF
Let $AB=s$ be the side of a regular decagon inscribed in the circle $\left( {O,R} \right)$ .
The central angle of the regular decagon is $\angle AOB = \frac{{360^\circ }}{{10}} = 36^\circ$ , hence $\angle OAB = \angle OBA = 72^\circ$ .

Draw $AC$ , the angle bisector of $\angle OAB$ .
We notice that both $\vartriangle COA$ and $\vartriangle ABC$ are isosceles, since $\angle CAO = 36^\circ = \angle AOC$ and $\angle ACB = \angle CAO + 36^\circ = 72^\circ = \angle OBA$ .
Hence, $AB=AC=CO=s$ .

By angle bisector theorem on $\vartriangle OAB$ we get
$\frac{{AB}}{{AO}} = \frac{{CB}}{{CO}} \Leftrightarrow \frac{s}{R} = \frac{{R - s}}{s} \Leftrightarrow {s^2} + Rs - {R^2} = 0.$

That gives
$\begin{gathered} s = \frac{{ - R \pm \sqrt {5{R^2}} }}{2} = R \cdot \frac{{ - 1 \pm \sqrt 5 }}{2} \\ \mathop \Rightarrow \limits^{s > 0} s = R \cdot \frac{{ - 1 + \sqrt 5 }}{2} \Leftrightarrow R = s \cdot \frac{2}{{ - 1 + \sqrt 5 }} = s \cdot \frac{{1 + \sqrt 5 }}{2} \Leftrightarrow R = s \cdot \varphi \\ \end{gathered}$

Step 5 :
Let ${R_1}$ , ${R_2}$ be the radii of circles $c_1$ and $c_2$ respectively. Label ${s_1} = MN = \sqrt 2$ and ${s_2} = CD$ .
By the theorem of step 4, we have ${R_1} = {s_1} \cdot \varphi \Rightarrow {R_1} = \varphi \cdot \sqrt 2. \ \ \ \ \ (1)$

By Pythagoras’ theorem on $\vartriangle ABN$ , $A{B^2} = N{A^2} + N{B^2} \Rightarrow {\left( {2{R_1}} \right)^2} = 2{R_2}^2 \Rightarrow {R_2} = {R_1} \cdot \sqrt 2 \mathop \Rightarrow \limits^{\left( 1 \right)} {R_2} = \varphi \cdot {\sqrt 2 ^2} \Rightarrow {R_2} = 2\varphi$ .

Step 6 :
By similarity of regular decagons $\frac{{{s_2}}}{{{s_1}}} = \frac{{{R_2}}}{{{R_1}}} \Rightarrow {s_2} = {s_1} \cdot \frac{{{R_2}}}{{{R_1}}} \Rightarrow {s_2} = \sqrt 2 \cdot \sqrt 2 \Rightarrow {s_2} = 2$ .

By Pythagoras’ theorem on $\vartriangle BCD$ ,
$\begin{gathered} B{D^2} = B{C^2} + C{D^2} \Rightarrow B{C^2} = B{D^2} - C{D^2} \\ \Rightarrow B{C^2} = {\left( {2{R_2}} \right)^2} - {2^2} = {\left( {4\varphi } \right)^2} - 4 = 16{\varphi ^2} - 4 \\ \end{gathered}$ .

Since ${\varphi ^2} = \varphi + 1$ , we get $B{C^2} = 16\left( {\varphi + 1} \right) - 4 = 16\varphi + 12 \Rightarrow BC = 2\sqrt {4\varphi + 3}$ .

Now we have
$\begin{gathered} {\text{area of }} \vartriangle ABCD = {\text{area of }} \vartriangle ABD + {\text{area of }} \vartriangle BCD \\ = \frac{1}{2}AB \cdot AD + \frac{1}{2}BC \cdot CD \\ = \frac{1}{2}{\left( 2R_1\right)}^2 + \frac{1}{2}BC \cdot 2 \\ = \frac{1}{2} \cdot 8{\varphi ^2} + \frac{1}{2} \cdot 2\sqrt {4\varphi + 3} \cdot 2 \\ = 4\left( {\varphi + 1} \right) + 2\sqrt {4\varphi + 3} \\ = 4\varphi + 4 + 2\sqrt {4\varphi + 3} \\ = \left( {4\varphi + 3} \right) + 2\sqrt {4\varphi + 3} + 1 \\ = {\left( {\sqrt {4\varphi + 3} + 1} \right)^2} \\ \end{gathered}$ .

For the answer, $a=4$ , $b=3$ , $c=1$ , thus $a+b+c=\boxed{8}$ .