$45°$ is everywhere

Since $\angle CAB = \angle CBA$ , it follows that $AC = BC$ . If we take $\triangle NBC$ and rotate it clockwise by $90°$ about point $C$ , side $BC$ will completely overlap side $AC$ , point $B$ will end up in point $A$ , and point $N$ will end up in a point marked as $E$ . Now, we have that:

$\angle ACB = 90° \rightarrow \angle CAM = \angle CBN = \angle EAC = 45° \rightarrow \angle EAM = \angle EAC + \angle CAM = 90°$ .

$\triangle EAM$ is a right triangle, and by Pythagorean Theorem: $EM = \sqrt{EA^2 + AM^2} = 5$ .

Now, $\triangle CEM$ and $\triangle CMN$ are congruent, because:

• $\angle ECM = \angle ECA + \angle ACM = \angle NCB + \angle ACM = \angle ACB - \angle MCN = 45° = \angle MCN$
• $EC = CN$
• side $CM$ is common to both of these triangles.

From this, it must be that: $MN = EM = 5$ . So:

• $AB = AM + MN + NB = 3 + 5 + 4 = 12$

• $AC = BC = \dfrac{AB}{\sqrt{2}}$

• $P = \dfrac{AC^2}{2} = \dfrac{AB^2}{4} = \boxed{36}$

Since $\angle ACB = 90^\circ$ then $\angle A = \angle B = 45^\circ = \angle MCN$ . Let $AC=a$ . Since $\angle A = \angle B$ then $AC=BC=a$ . Let $\angle ACM = \theta$ , then $\angle CMA = 180^\circ - 45^\circ - \theta = 135^\circ - \theta$ . By sine rule : $\dfrac {AC}{AM} = \dfrac a3 = \dfrac {\sin \angle CMA}{\sin \angle ACM} = \dfrac {\sin (135^\circ - \theta)}{\sin \theta}$

And $\angle NCB = 90^\circ - 45^\circ - \theta = 45^\circ - \theta$ and $\angle CNB = 180^\circ - 45^\circ - (45^\circ - \theta) = 90^\circ + \theta$ . By sine rule again: $\dfrac {BC}{NB} = \dfrac a4 = \dfrac {\sin (90^\circ + \theta)}{\sin (45^\circ - \theta)}$ .

Then we have:

$\begin{aligned} a = \frac {3\sin (135^\circ - \theta)}{\sin \theta} & = \frac {4\sin (90^\circ + \theta)}{\sin (45^\circ - \theta)} \\ \frac {3\left(\frac 1{\sqrt 2}\cos \theta + \frac 1{\sqrt 2}\sin \theta\right)}{\sin \theta} & = \frac {4\cos \theta}{\frac 1{\sqrt 2}\cos \theta - \frac 1{\sqrt 2}\sin \theta} \\ \frac 32 (\cos^2 \theta - \sin^2 \theta) & = 4 \sin \theta \cos \theta \\ \frac 32 \cos 2\theta & = 2 \sin 2 \theta \\ \implies \tan 2\theta & = \frac 34 \\ \frac {2 \tan \theta}{1-\tan^2 \theta} & = \frac 34 \\ 3\tan^2 \theta +8\tan \theta - 3 & = 0 \\ (3\tan \theta -1)(\tan \theta + 3) & = 0 \\ \implies \tan \theta & = \frac 13 \quad \quad \small \color{#3D99F6} \text{As } \theta < 90^\circ \end{aligned}$

Now we have:

$\begin{aligned} a & = \frac {4\sin (90^\circ + \theta)}{\sin (45^\circ - \theta)} = \frac {4\cos \theta}{\frac 1{\sqrt 2}\cos \theta - \frac 1{\sqrt 2}\sin \theta} = \frac {4\sqrt 2}{1-\tan \theta} = \frac {4\sqrt 2}{1-\frac 13} = 6\sqrt 2 \end{aligned}$

Therefore, the area of $\triangle ABC$ : $A = \dfrac {a^2}2 = \boxed{36}$ .

We reflect $A$ to $MC$ line. If we get the $Q$ point, then

$QC=AC, \angle QCM=\angle ACM.$

We reflect $B$ to $NC$ line. If we get the $R$ point, then

$RC=BC, \angle RCN=\angle BCN$ .

If $\angle ACM=\alpha$ , then $\angle QCM=\alpha$ , and if $\angle BCN=\beta$ , then $\angle RCN=\beta$ .

If $Q$ and $R$ is not the same point, then $\angle MCN=\alpha+\beta+\angle QCR$ or $\angle MCN=\alpha+\beta-\angle QCR$ . (It depends on that $Q$ is near to $B$ and $R$ is near to $A$ , or inversely.) We know that $\angle MCN=45°$ , so $\angle ACB=90°-\angle QCR$ or $\angle ACB=90°+\angle QCR$ . But $\angle ACB=90°$ , so it is impossible unless $Q$ and $R$ is the same point. (Let $P$ the point which was $R$ and $Q$ .) By the reflection we get:

$\angle MPC=\angle MAC=45°$

$\angle NPC=\angle NBC=45°$

$\angle MPN=\angle MPC+\angle NPC=\angle MAC+\angle NBC=90°$

So the $NMP$ triangle is right-angled. Using Pythagoras-theorem: $MN^2=PM^2+PN^2.$

Since $PM=AM$ and $PN=NB$ , $MN^2=AM^2+NB^2$ . So $MN=\sqrt{MN^2}=\sqrt{3^2+4^2}=5$ .

From that $AB=3+4+5=12$ and

$[ABC]=\dfrac{AB*\dfrac{AB}{2}}{2}=\boxed{36}$

Note that $\angle ACN = \angle ACM + \angle MCN = \angle ACM + \angle CAM = \angle CMB$ and $\angle A = \angle B$ so $\triangle ACN \sim \triangle BMC$ . Since $\triangle ACB$ is $45^\circ -45^\circ-90^\circ$ , we have $BC =AC$ , $AB = AC \sqrt{2}$ and $MN = AB - AM -NB = AC\sqrt {2} - 7$ , from which it follows that $AN = AM + MN = AC\sqrt{2} - 4$ and $MB = MN + NB = AC\sqrt{2} - 3$ . From the stated similarity, $\frac {AC}{BM} = \frac{AN}{BC}$ . Substituting into this last proportion yields $\frac {AC}{AC\sqrt{2}-3}=\frac{AC\sqrt{2}-4}{AC}$ . Now solve for $AC$ to obtain $AC = \sqrt{2}$ , an impossibility, or $AC = 6\sqrt{2}$ . This last value implies the requested area is $\frac{AC^2}{2 }= 36$ .

Let $AM=p$ , $MN=r$ , and $NB=q$ . We will first show the interesting fact that $p^2+q^2=r^2$ .

Draw $\overline{NP}$ so that $NP=NB$ and $\angle CNP=\angle CNB$ . Draw $\overline{CP}$ .

$\triangle CNB\cong\triangle CNP$ by SAS, so $CP=CB$ and $\angle PCN=\angle BCN$ . Now draw $\overline{PM}$ .

Now $\angle ACM=\angle ACB-\angle MCN-\angle NCB=90^\circ-45^\circ-\angle NCB=45^\circ-\angle NCB$ $=\angle MCN-\angle PCN=\angle PCM$ .

So $\angle ACM=\angle PCM$ and $CP=CB=CA$ , so $\triangle ACM\cong\triangle PCM$ by SAS. Thus, $PM=AM=p$ .

Triangle $MPN$ is a right triangle because $\angle CPM=\angle CAM=45^\circ$ and $\angle CPN=\angle CBN=45^\circ$ .

So $PM=p$ , $NP=q$ , and $MN=r$ , and by the Pythagorean Theorem, $p^2+q^2=r^2$ .

Now, getting back to the original problem, $p=3$ and $q=4$ , so $r=5$ . Therefore, $AB=12$ which implies that the area of $\triangle ABC=36$ .

Let $MN=x , AC=BC=y$

Pythagorean theorem on $ΔABC$

$(x+7)^2=y^2+y^2\Leftrightarrow x^2+14x+49=2y^2$ (1)

.

$\Delta ANC\approx \Delta BCM$

$\frac{AC}{BM}=\frac{AN}{BC}\Leftrightarrow \frac{y}{x+4}=\frac{x+3}{y}\Leftrightarrow y^2=x^2+7x+12$ (2)

.

From (1) and (2) $x = 5$ and $y^2=72$

The area of $ΔABC$ is : $[ABC]=\frac{y^2}{2}=\frac{72}{2}= 36$

Let $O$ be the midpoint of side $\bar{AB}$ , then it is obvious it's the perpendicular bisector, hence, $\bar{AO}$ = $\bar{BO}$ = $\bar{CO}$ = x, then since $\angle MCN$ is 45,
$tan^{-1}(\frac{x-3}{x}$ ) + $tan^{-1}(\frac{x-4}{x}$ ) = $\frac{\pi}{4}$

then using $tan(a+b) = \frac{tan(a)+tan(b)}{1-tan(a)tan(b)}$ ,

$tan( tan^{-1}(\frac{x-3}{x}$ ) + $tan^{-1}(\frac{x-4}{x}$ )) = tan( $\frac{\pi}{4}$ )

$\frac{ tan(tan^{-1}(\frac{x-3}{x} )) + tan(tan^{-1}( \frac{x-4}{x} )) }{1- tan(tan^{-1}( \frac{x-3}{x}))tan(tan^{-1}(\frac{x-4}{x} )) }$ =1 ,

resulting in a quadratic equation: \

$\frac{ 1 - \frac{3}{x} + 1 - \frac{4}{x} } { 1- (1-\frac{3}{x})(1-\frac{4}{x}) } =1$

whose solution is $x = 6$ , hence the area is $A=\frac{1}{2}*2x*x = x^2 = 6^2 = 36$