\( \newcommand{\bx}{\boldsymbol{x}} \newcommand{\by}{\boldsymbol{y}} \newcommand{\bp}{\boldsymbol{p}} \newcommand{\bq}{\boldsymbol{q}} \newcommand{\br}{\boldsymbol{r}} \renewcommand{\bk}{\boldsymbol{k}} \renewcommand{\ba}{\boldsymbol{a}} \renewcommand{\bb}{\boldsymbol{b}} \newcommand{\bc}{\boldsymbol{c}} \newcommand{\bh}{\boldsymbol{h}} \newcommand{\bA}{\boldsymbol{A}} \newcommand{\bB}{\boldsymbol{B}} \newcommand{\bC}{\boldsymbol{C}} \newcommand{\definedas}{\equiv} \newcommand{\half}{\frac{1}{2}} \newcommand{\bsigma}{\boldsymbol \sigma} \newcommand{\rectangle}{{% \ooalign{$\sqsubset\mkern3mu$\cr$\mkern3mu\sqsupset$\cr}% }} \newcommand{\Rectangle}{\sqsubset\!\sqsupset} \newcommand{\abs}[1]{\left|\,#1\,\right|} \)

Triangles

Here are presented papers on the triangle, but not using geometric algebra. Those papers are elsewhere
on this website.

The Angle-Bisector Theorem:
This paper demonstrates a possibly novel approach to the proof of the ancient Angle-Bisector Theorem.

The Right Triangle Diameter Theorem:
Let the vertex at which is the right angle of a right triangle be called the `apex' of the triangle, just to give
it a convenient label. Let the right triangle have its hypotenuse exactly over a diameter of a circle of radius $r$.
Then, of course, the apex of the triangle is either inside the circle, outside the circle, or on the circle. I intend
to prove that it must be on the circle.


Setup: Let points B,C be points on the circle, being connected by a
diameter of the circle, whose center is at point 0. Point A is the apex
point of the triangle. Therefore, as explained above, the non-apex points
of the triangle are on points B,C, and the angle at A is a right angle.