How to find area $\triangle ABC$, where $AC=BD, CE=2, ED=1, AE=4$ and $\angle CAE=2 \angle DAB$ In a triangle $\triangle ABC$, $AC=BD, CE=2, ED=1, AE=4$ and $\angle CAE=2  \angle DAB$. Find area of $\triangle ABC$.

This is what it looks like, but how to justify the red ones or maybe there is another way to solve this problem?


Take a Cartesian coordinate system where $\vec d=(0,0)$,
$\vec e=(1,0)$, $\vec c=(3,0)$ and $\vec b=(-b,0)$ with $b>0$,
$\vec a=(a_x,a_y)$. So $CE=2$ and $ED=1$ are satisfied.
$$ AC=BD \to
b^2=(3-a_x)^2+a_y^2
% \to
% b^2=9+a_x^2-6a_x+a_y^2
\to
b^2-9+6a_x=a_x^2+a_y^2
$$
$$ AE=4 \to
(a_x-1)^2+a_y^2=4^2.
\to
a_x^2-2a_x+1+a_y^2=16
\to
a_x^2+a_y^2=15+2a_x.
$$
$$
b^2=24-4a_x.
$$
$$
b=2\sqrt{6-a_x}.
$$
Vectors $\vec{AB}=(-b-a_x,-a_y)$,
$\vec{AD}=(-a_x,-a_y)$ and the dot product is
$$
\vec{AB}\cdot \vec{AD}=\cos\alpha |\vec{AB}||\vec{AD}|.
$$
$$
-a_x(-b-a_x)+a_y^2=\cos\alpha \sqrt{(-b-a_x)^2+a_y^2}\sqrt{a_x^2+a_y^2}.
$$
$$
+a_x b+15+2a_x=\cos\alpha \sqrt{b^2+2ba_x +15+2a_x}\sqrt{15+2a_x}.
$$
$$
2a_x \sqrt{6-a_x}+15+2a_x=\cos\alpha \sqrt{2ba_x +39-2a_x}\sqrt{15+2a_x}.
$$
Vector $\vec{AE}=(1-a_x,-a_y)$ and
$\vec{AC}=(3-a_x,-a_y)$, so the dot product is
$$
\vec{AE}\cdot \vec{AC}=\cos(2\alpha) |\vec{AE}||\vec{AC}|.
$$
$$
(1-a_x)(3-a_x)+a_y^2=\cos(2\alpha) \sqrt{(1-a_x)^2+a_y^2}\sqrt{(3-a_x)^2+a_y^2}
=\cos(2\alpha) 4 b
$$
$$
3-4a_x +15+2a_x
=\cos(2\alpha) 4 b
$$
$$
18-2a_x
=\cos(2\alpha) 4 b
$$
$$
9-a_x
=\cos(2\alpha) 2\cdot 2\sqrt{6-a_x}
=4\sqrt{6-a_x}(2\cos^2\alpha-1)
$$
Numerical solution of these 2 coupled equations for $\cos \alpha$ and $a_x$ converges
to $AC=BD=b=3$, $\alpha\approx 14.4775^\circ$, $a_x=3\frac34$, $a_y=3\sqrt{15}/4$. The area is half the product of baseline and altitude, $(b+3)a_y/2$.
