# A Computer Algebra Aided Proof in Plane Geometry

Given $\Delta ABC$ and two squares $ABEF, ACGH$ in Fig. 1. The squares are sitting on the two sides of $\Delta ABC, AB$ and $AC$, respectively. Both squares are oriented away from the interior of $\Delta ABC$. $\Delta BCP$ is an isosceles right triangle. $P$ is on the same side of $A$. Prove that points $E, P$ and $G$ lie on the same line.

Fig. 1

Introducing rectangular coordinates show in Fig. 2:

Fig. 2

We observe that $y>0\quad\quad\quad(1)$ $x_1 < -a\quad\quad(2)$ $x_3 > a\quad\quad\quad(3)$ $CG=CA\implies (x-a)^2+y_3^2 = (x-a)^2 +y^2\quad\quad(4)$ $AB=AE\implies (x+a)^2+y^2=(x_1+a)^2+y_1^2\quad\quad(5)$ $CG\perp CA \implies y_3 y = -(x-a)(x_3-a)\quad\quad\quad(6)$ $BE\perp AB \implies y_1y = -(x_1+a)(x+a)\quad\quad\quad(7)$

Fig. 3

Solving system of equations (4), (5), (6), (7), we obtain four set of solutions: $x_1=-y-a, y_1=x+a, x_3=y+a, y_3=a-x\quad\quad(8)$ $x_1=y-a, y_1=-x-a, x_3=y+a, y_3=a-x\quad\quad(9)$ $x_1=-y-a, y_1=x+a, x_3=a-y, y_3=x-a\quad\quad(10)$ $x_1=y-a, y_1=-x-a, x_3=a-y, y_3=x-a\quad\quad(11)$

Among them, only (10) truly represents the coordinates in Fig. 2.

Fig. 4

By Heron’s formula (see “Had Heron Known Analytic Geometry…“), the area of triangle with vortex $(-y-a, x+a), (0, a), (y+a, a-x)$ is $\frac{1}{2}\begin{vmatrix} -y-a & x+a & 1 \\ 0 & a &1 \\ y+a & a-x & 1\end{vmatrix}$.

From Fig. 4, we see that it is zero.

Therefore, $E, P, G$ lie on the same line.

The reason we do not consider (9), (10), (11) is due to the fact that

(9) contradicts (2) since $y>0, a>0 \implies x_1=y-a=-a+y>-a$.

And,

by (1), (10) and (11) indicate $x_3=a-y which contradicts (3).

Exercise-1 Prove “ $E, P, G$ lie on the same line” with complex numbers (hint: see “Treasure Hunt with Complex Numbers“).