Cool bisector

First we have to find the measurement of $BD$

Let $AB=a, BC=b, AC=c, DC=x$ and $BD=y$

By angle bisector theorem

$\frac{a}{b}=\frac{c-x}{x} \Rightarrow$ $ax=bc-bx$ so $x=\frac{bc}{a+b}$

Then by law of sines

$\frac{\frac{bc}{a+b}}{ \sin 45°}=\frac{y}{ \sin \angle C}$

$y=\frac{\frac{bc}{a+b}}{ \sin 45°}\times \sin \angle C$

$y=\frac{\frac{bc}{a+b}}{ \sin 45°}\times \frac{a}{c}$

$\boxed{y=\frac{ab}{(a+b) \sin45°}}$ this is the bisector's length of in any rectangle triangle.

Second we have to find the area

Area of $\triangle EAB=\frac{1}{2} ay \sin (180-A)$ but $\sin (180-A)= \sin A$

Area of $\triangle EAB=\frac{1}{2} \frac{ab}{(a+b) \sin45°} \sin (A)$

Area of $\triangle EAB=\frac{1}{2} \times \frac{12}{(7) \sin45°} \times \frac{4}{5}$

Area of $\triangle EAB=\boxed{\frac{72 \sqrt{2}}{35} \approx 2.91}$

All I did was $BF \perp AC \quad and \quad DG \perp BC$

Thus $[ EAB ] = \dfrac{(AE)(BF)}{2} = \dfrac{\left(\dfrac{12\sqrt{2}}{7}\right)\left(\dfrac{12}{5}\right)}{2} = \dfrac{72\sqrt{2}}{35} \approx 2.909 \approx 2.91$

I have no idea how sines work so I used this to get 2.88:

I assumed angle $BDA$ was a right angle. With the pythagorean theorem, we get $a^2+b^2=c^2$ Substituting... $(3)^2+(4)^2=c^2$ , $9+16=c^2$ , $25=c^2$ , $5=c$

Area of $ΔABC = \frac{(3)(4)}{2} = 6$

Let's call the length of line $BD$ $'x'$ . Then $\frac{5x}{2}=6$ ... $x=(12/5)=2.4$

Now, since we know that $EA=BD$ , $EA=2.4$ . I'll call the length of line $EA$ $'y'$ .

And finally, we find the area of $ΔEBC + ΔABC$ : $\frac{(5+y)(2.4)}{2} = \frac{(5+(2.4))(2.4)}{2} = \frac{(7.4)(2.4)}{2} = \frac{17.76}{2}=8.88$

Subtract the area of $ΔABC$ for Δ $EAB$ :

$8.88-6=\boxed{2.88}$

$let ~x=AD ~ and ~ y=DC.~~~\alpha=\angle BDA,~~~BF\perp AC ,~ F~on~AC. \\ \sqrt{3^2+4^2}=\color{#3D99F6}{5=AC=x+y.}\\ Bisector~ Theorem:- \dfrac x 3=\dfrac y 4=\dfrac {5 - x}4, ~~\therefore x=\dfrac{15} 7,~~and~~y=5 - x=\dfrac {20} 7.\\ \color{#3D99F6}{BF= 3*SinBAC=3*\dfrac 4 5=\dfrac{12} 5}.\\ \color{#3D99F6}{AE=BD=\sqrt{3*4 - x*y}=\sqrt{12 - \dfrac{300}{49} } }\\ Area~\Delta EAB=\frac 1 2 *AE*BF = \Large~~~\color{#D61F06}{2.909}$

$let ~x=AD ~ and ~ y=DC.~~~\alpha=\angle BDA \\ \sqrt{3^2+4^2}=\color{#3D99F6}{5=AC=x+y.}\\ App. ~Sine~ Rule~:-~~\dfrac x {Sin45}=\dfrac 3 {Sin\alpha},~~~\dfrac y {Sin45}=\dfrac 4 {Sin(180 - \alpha)}.\\ Adding~the ~ two ~ equations ~ Sin\alpha=\dfrac 7 {5*\sqrt 2}.\\ App. ~Sine~ Rule~:- ~ \dfrac x {Sin45}=\dfrac 3 {Sin\alpha},~~~\therefore~ x=\dfrac{15} 7,~~and~~y=5 - x=\dfrac {20} 7.\\$ $Bisector~ Theorem:- \color{#3D99F6}{ AE=BD=\sqrt{3*4 - x*y}=\sqrt{12 - \dfrac{300}{49}} }.\\$ $Let~BF\perp AC ~~~~~~~~~~Note:- BF\perp AC , BA:AF:BF::5:3:4 \\ In~ rt.~\angle d~\Delta ABC,~ BF\perp AC,~\implies BF\perp~EA,=\dfrac 3 5 *4=\dfrac{12} 5.\\ Area~\Delta EAB=\frac 1 2 *AE*BF = \Large~~~\color{#D61F06}{2.909}$

I used the angle bisector theorem and triangle similarity. I extended line BA and drew a line from point E such that it is perpendicular to extended line BA. Let the point of intersection of those two aforementioned lines be X. To find the area, we are already given the "base" of triangle EAB, which is 3, so we just need the value of line EX. We already know that via pythagorean theorem, AC= 5.Use $\frac{a}{x} = \frac{b}{y}$ and the previous information to obtain the values of AD and DC (system of equations).Then plug into $e^{2}=ab-xy$ to get e (which is DB). Since DB=EA, use triangle similarity to obtain the value of EX. Then use area of triangle to get the final answer.