Prove that $S = 8S_1-4S_2$ in the geometrical figure

Let squares be constructed on the sides $BC,CA,AB$ of triangle $ABC$, all to the outside of the triangle, and let $A_1,B_1,C_1$ be their centers. Starting from the triangle $A_1B_1C_1$ one analogously obtains a triangle $A_2B_2C_2$. If $S,S_1,S_2$ denote the areas of triangles $ABC,A_1B_1C_1,A_2B_2C_2,$ respectively, prove that $S = 8S_1-4S_2$

Edits: I made several edits to this post and what is on this post now is the most correct version. If anyone wants to provide a geometrical solution to this question that isn't computational, that would be very helpful.

Attempt:

Note: Throughout this solution I will be using the fact that $\vec{a}+\vec{b}+\vec{c} = \vec{0}$. In other words, $a_1 + b_1+c_1 = 0$ and $b_2+c_2 = 0$.

If we let the sides of the triangle $ABC$ be $\vec{a} = \begin{bmatrix} a_1 \\ 0 \end{bmatrix},\vec{b} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix},\vec{c} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$ oriented counterclockwise, then $$S = \frac{1}{2} \left| \left|\vec{a} \times \vec{b}\right| \right| = \frac{1}{2} \left | \left|\begin{bmatrix} a_1 \\ 0 \\ 0 \end{bmatrix} \times \begin{bmatrix} b_1 \\ b_2 \\ 0 \end{bmatrix} \right | \right | = \frac{1}{2} \left|a_1b_2 \right|.$$

Now notice that by the picture above that $$\vec{KL} = \dfrac{1}{2}R(90^{\circ})\vec{b}+\frac{1}{2}\vec{c}+\frac{1}{2}R(-90^{\circ})\vec{a} = \dfrac{1}{2} \left(\begin{bmatrix} -b_2 \\ b_1 \end{bmatrix}+\begin{bmatrix} c_1 \\ c_2 \end{bmatrix}+\begin{bmatrix} 0 \\ -a_1 \end{bmatrix} \right) = \dfrac{1}{2} \begin{bmatrix} c_1-b_2 \\ b_1+c_2-a_1 \end{bmatrix} = \dfrac{1}{2}\begin{bmatrix} c_1-b_2 \\ -c_1-b_2-2a_1 \end{bmatrix}$$ and $$\vec{LJ} = \dfrac{1}{2}R(-90^{\circ})\vec{c}+\frac{1}{2}\vec{b}+\frac{1}{2}R(90^{\circ}) \vec{a} = \dfrac{1}{2} \left(\begin{bmatrix} c_2 \\ -c_1 \end{bmatrix}+\begin{bmatrix} b_1 \\ b_2 \end{bmatrix}+\begin{bmatrix} 0 \\ a_1 \end{bmatrix} \right) = \dfrac{1}{2} \begin{bmatrix} b_1+c_2 \\ b_2-c_1+a_1 \end{bmatrix} = \dfrac{1}{2} \begin{bmatrix} -a_1-c_1-b_2 \\ b_2-c_1+a_1 \end{bmatrix}.$$

Note: We will need this result for later that $\vec{JK} = \dfrac{1}{2} \left(R(90^{\circ} \cdot) \vec{c}+\vec{a}+R(-90^{\circ}) \cdot \vec{b}\right) = \dfrac{1}{2}\left(\begin{bmatrix} -c_2 \\ c_1 \end{bmatrix}+\begin{bmatrix} a_1 \\ 0 \end{bmatrix}+\begin{bmatrix} b_2 \\ -b_1 \end{bmatrix} \right) = \dfrac{1}{2} \begin{bmatrix} 2b_2+a_1 \\ c_1-b_1 \end{bmatrix}.$

It follows that $$S_1 = \frac{1}{2} \left| \left|\vec{KL} \times \vec{LJ}\right| \right| = \dfrac{1}{8} \left(\left| \left| \begin{bmatrix} c_1-b_2 \\ -c_1-b_2-2a_1 \end{bmatrix} \times \begin{bmatrix} -a_1-c_1-b_2 \\ b_2-c_1+a_1 \end{bmatrix} \right| \right| \right)= \dfrac{1}{8}\left|-2 a_1^2-4 a_1 b_2-2 b_2^2-2 a_1 c_1-2 c_1^2\right|.$$

Finally, to find $S_2$ we apply the same procedure. First notice that $\triangle{KLJ}$ is oriented clockwise. Then that $\vec{WT} = \dfrac{1}{2}R(-90^{\circ}) \cdot \vec{KL} -\dfrac{1}{2}\vec{JK}+\dfrac{1}{2}R(90^{\circ}) \cdot \vec{LJ}$ and $\vec{TQ} = \dfrac{1}{2}R(-90^{\circ}) \cdot \vec{LJ}-\dfrac{1}{2}\vec{KL}+ \dfrac{1}{2}R(90^{\circ}) \cdot \vec{JK}$. Thus, $$\vec{WT} = \dfrac{1}{2} \left(\dfrac{1}{2} \begin{bmatrix}-c_1-b_2-2a_1 \\ -(c_1-b_2) \end{bmatrix} - \dfrac{1}{2} \begin{bmatrix} 2b_2+a_1 \\ 2c_1+a_1 \end{bmatrix}+\dfrac{1}{2} \begin{bmatrix}-(b_2-c_1+a_1) \\ -a_1-c_1-b_2 \end{bmatrix} \right) = \dfrac{1}{4} \begin{bmatrix} -4a_1-4b_2 \\ -2a_1-4c_1 \end{bmatrix}$$ and $$\vec{TQ} = \dfrac{1}{2}\left(\dfrac{1}{2} \begin{bmatrix}b_2-c_1+a_1 \\-(-a_1-c_1-b_2) \end{bmatrix}- \dfrac{1}{2} \begin{bmatrix} c_1-b_2 \\ -c_1-b_2-2a_1 \end{bmatrix} +\dfrac{1}{2} \begin{bmatrix}-(2c_1+a_1) \\ 2b_2+a_1 \end{bmatrix}\right) = \dfrac{1}{4} \begin{bmatrix} -4c_1+2b_2 \\ 4 a_1+4 b_2+2 c_1 \end{bmatrix}.$$

Finally, $$S_2 = \dfrac{1}{2} \left| \left|\vec{WT} \times \vec{TQ}\right| \right| = \dfrac{1}{32} \left |-16 a_1^2-28 a_1 b_2-16 b_2^2-16 a_1 c_1-16 c_1^2 \right |.$$

To verify the equation, we see that $S = 8S_1-4S_2$ is equivalent to $\dfrac{1}{2}\left|a_1b_2 \right| = 8 \left(\dfrac{1}{8} \left |-2 a_1^2-4 a_1 b_2-2 b_2^2-2 a_1 c_1-2 c_1^2 \right| \right)-4 \left(\dfrac{1}{32} \left |-16 a_1^2-28 a_1 b_2-16 b_2^2-16 a_1 c_1-16 c_1^2 \right | \right) = \left |-2 a_1^2-4 a_1 b_2-2 b_2^2-2 a_1 c_1-2 c_1^2 \right|-\dfrac{1}{8} \left|-16 a_1^2-28 a_1 b_2-16 b_2^2-16 a_1 c_1-16 c_1^2\right|.$

As you can see, the last equality looks right but how do I prove it is true for a triangle?

Let me try. Denote that $AB=c$, $BC=a$, $CA=b$. One then has $$S_1 = S + a^2+b^2+c^2 - \left(\sum \frac{1}{2}a\sqrt{2}b\sqrt{2} \sin(C+90^\circ)\right) = S+ a^2 + b^2 + c^2 -\left(\sum ab\cos C\right)$$

$$=S+a^2+b^2+c^2 - \sum \frac{a^2+b^2-c^2}{2} = S+ \frac{1}{2}(a^2+b^2+c^2).$$

In the same way, one has $$S_2 = S_1 + \frac{1}{2}(A_1B_1^2+B_1C_1^2+C_1A_1^2).$$

Note that $A_1B_1^2 = 2a^2+2b^2-2(2ab)\cos(C+90^\circ) = 2a^2+2b^2+4ab\sin C = 2a^2+2b^2 + 8S$.

Thus, $$S_2 = S_1 + \frac{1}{2}(4(a^2+b^2+c^2) + 24S)$$

So $$S_2 = S_1 + 2(a^2+b^2+c^2) + 12S = S_1 + 4(S_1-S)+12S = 5S_1 + 8S.$$

EDIT: I think the case of one angle is more than $90^\circ$ is the same.