# A Complex Number Aided Proof in Plane Geometry

The proof in my post “A Computer Algebra Aided Proof in Plane Geometry” can be significantly simplified if complex numbers are used.

Consider complex numbers $v_1, v_2, w_1, w_2, p$ in FIg. 1.

Fig. 1

Since

$v_1=p-(-a)= p+a \overset{p=x+\bold{i}y}{=} x+\bold{i}y+a \implies v_1 = x+a+\bold{i}y,\quad\quad\quad(1)$

we have,

$\bold{i}v_1 = w_1-(-a) \implies w_1=-a+\bold{i} v_1 \overset{(1)}{=} -a +\bold{i}(x+a+\bold{i}y)=-a-y+\bold{i}(x+a).$

i.e.,

$\begin{cases} x_1 = -a-y \\ y_1=x+a\end{cases}\quad\quad\quad(*)$

Similarly,

$v_2 = p-a = (x+\bold{i}y)-a=x-a+\bold{i}y\quad\quad\quad(2)$

and

$-\bold{i}v_2=w_2-a\implies w_2 = a + (-\bold{i} v_2) \overset{(2)}{=} a + (-\bold{i}(x-a+\bold{i}y)) = a+y+\bold{i}(a-x)$

give

$\begin{cases} x_3 = a+y \\ y_3 = a-x \end{cases}\quad\quad\quad(**)$

Let $k_1, k_3$ denote the slope of line connecting $(0, a)$ and $(x_1, y_1), (x_3, y_3)$ respectively.

From

$k_1=\frac{y_1-a}{x_1-0} \overset{(*)}{=} \frac{x+a-a}{-a-y} = \frac{-x}{a+y},$

$k_3=\frac{y_3-a}{x_3-0} \overset{(**)}{=} \frac{a-x-a}{a+y-0} = \frac{-x}{a+y},$

we see that

$k_1 = k_3\implies (x_1, y_1), (0, a), (x_3, y_3)$ lie on the same line.

Notice that $a+y \ne 0$ since $y>0, a>0$.