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 $y>0, a>0$ .

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