Product of slopes is -1 iff perpendicular proof from first principles

Here's an elementary, trig-free proof.

Suppose the slopes $t_1,t_2$ of the lines $L_1,L_2$, respectively, are both defined (real numbers) and the lines intersect.

Let $p$ be their intersection. Then $q=p+(1,t_1)\in L_1$ and $r=p+(1,t_2)\in L_2$.

Now $L_1$ is perpendicular to $L_2$ if and only if the triangle $pqr$ has a right angle at $p$. By Pythagoras' theorem, this is equivalent to $\|p-q\|^2+\|p-r\|^2=\|q-r\|^2\iff 1+t_1^2+1+t_2^2=(t_1-t_2)^2\iff t_1t_2=-1$.

I'm not sure what geometric properties you're allowed to use as yet, but here's an attempt at a purely-geometric proof (trig-free).

Let's suppose, for convenience, that the point of intersection of the two lines is not on the $x$-axis and that neither line is horizontal or vertical. Call the point of intersection of each line with the $x$-axis $A$ and $B$ and the intersection point of the two lines $C$. Call the intersection of the vertical line through $C$ with the $x$-axis $D$. Looking at $\triangle ADC$, $\frac{DC}{AD}$ is the absolute value of the slope of the line that contains $A$ and $C$; similarly, $\frac{DC}{BD}$ is the absolute value of the slope of the line that contains $B$ and $C$.

If the lines are perpendicular, than $\angle ACB$ is a right angle, so $\triangle ABC$ is a right triangle, and $CD$ is the geometric mean of $AD$ and $BD$, so $AD\cdot BD=CD^2$, from which $\frac{DC}{AD}\cdot\frac{DC}{BD}=1$, so the product of the absolute values of the slopes is $1$. Since the slopes clearly have opposite signs, their product is $-1$.

If the product of the slopes is $-1$, then $\frac{DC}{AD}\cdot\frac{DC}{BD}=1$ or $AD\cdot BD=CD^2$. If you reflect point $C$ over the $x$-axis to $C'$, $CD=C'D$ and $AD\cdot BD=CD\cdot C'D$, so by the power of a point theorem, $A$, $B$, $C$, and $C'$ lie on a circle and since $AB$ is the perpendicular bisector of $CC'$, $AB$ is a diameter of the circle, so $\angle ACB$ is a right angle. Hence, the lines are perpendicular.

Alternately, if the product of the slopes is $-1$, then $\frac{DC}{AD}\cdot\frac{DC}{BD}=1$ or $\frac{DC}{AD}=\frac{BD}{DC}$ and since $\angle ADC$ and $\angle BDC$ are both right angles, $\triangle ADC\sim\triangle CDB$, so $\angle DAC\cong\angle DCB$ and $\angle DCA\cong\angle DBC$. Now, looking at the measures of the interior angles of $\triangle ABC$, their sum must be $180°$, but $\angle ACB$ is the sum of two angles that are congruent to $\angle ABC$ and $\angle BAC$, so the measure of $\angle ACB$ must be half of $180°$, which is $90°$, so $\angle ACB$ is a right angle. Hence, the lines are perpendicular.

The slopes $t_1$ and $t_2$ are the tangents of the angles $\alpha_1$ and $\alpha_2$ the two lines make with the $x$-axis.

We have $\alpha_1 - \alpha_2 = \pi/2$

Therefore loosely $\tan{(\alpha_1 - \alpha_2)} = \tan{(\pi/2)} = \infty$

And $\displaystyle\tan{(\alpha_1 - \alpha_2)} = \frac{t_1-t_2}{1+t_1t_2}$ by the formula for the sum/difference of tangents.

So we see that $t_1t_2 = -1$