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Examples of Figuring Things Out in Triangle Geometry

A right triangle is half a rectangle

• This gives the area of a right triangle.
• This is the basis for trigonometry.

A triangle is the sum of two right triangles

Alternatively, this is dropping an altitude.

• This gives the area of a triangle.
• This shows that medians divide a triangle into six equal areas.
  • For the subtriangles which have the same half-side have the same altitude and thus the same area.
  • And further dropping altitudes we have that B+B+A=C+C+A, thus B=C.

Solving simple algebraic formulas

• Solving B+B+A=C+C+A yields B=C, and when showing that medians divide a triangle into six equal areas.
• The addition formula for trigonometric functions?

Symmetry of a triangle

What is true for one statement about an arbitrary triangle regarding its vertices and sides is true for all permutations. This simply shortens proofs.

• We can apply this in showing that medians divide a triangle into six equal areas.

Four times a right triangle is the difference of two squares.

• This proves the Pythagorean theorem but we also need to solve the equation {$2ab=(a+b)^2-b^2$}.

Universal Hyperbolic Geometry

In hyperbolic geometry, the quadrance {$q$} (between points) and spread {$S$} (between lines) are dual. Each is equal to the cross-ratio.

Quadrance Spread Duality Theorem. If {$a_1=A_1^{\perp}$} and {$a_2=A_2^{\perp}$}, then {$R(a_1,b_2;a_2,b_1)=q(a_1,a_2)=S(A_1,A_2)$}. Where {$a_1,a_2$} are points within a circle and {$b_1,b_2$} are points on their respective polars.

In the plane, two vectors indicate the same line if the area they span is zero.

{$x_1:y_1 = x_2:y_2 \iff \frac{x_1}{y_1} = \frac{x_2}{y_2} \iff x_1y_2 - x_2y_1 = 0$}

In three-dimensional space, two vectors indicate the same line if, in each of the three planes, the area they span is zero.

{$x_1:y_1:z_1 = x_2:y_2:z_2 \iff \frac{x_1}{y_1} = \frac{x_2}{y_2} \wedge \frac{x_1}{z_1} = \frac{x_2}{z_2} \wedge \frac{y_1}{z_1} = \frac{y_2}{z_2} \iff x_1y_2 - x_2y_1 = x_1z_2 - x_2z_1 = y_1z_2 - y_2z_1 = 0$}

Given point a = [x:y:z] and line L = (l:m:n).

a is on L (L passes through a) when {$xl+my=nz$} or when {$xl+my-nz=0$} or when {$a\circ L=0$}.

a and L are dual ({$a^{\perp}=L$} and {$a=L^{\perp}$}) when [x:y:z]=[l:m:n] or (x:y:z)=(l:m:n).

a is null when it lies on its dual line, thus when {$x^2+y^2=z^2$}, (when {$a\circ a^{\perp}=0$}), and likewise, L is null when it passes through its dual point.

Points {$a_1=[x_1:y_1:z_1]$} and {$a_2=[x_2:y_2:z_2]$} are perpendicular when {$x_1x_2+y_1y_2=z_1z_2$} or when {$x_1x_2+y_1y_2-z_1z_2=0$} or when {$a_1\circ a_2=0$}. Similarly with lines.

Overview

We have laws for:

• {$q_1q_2=S_1S_2$} Spread law.
• {$q_1q_2q_3$} Triple quad law.
• {$q_1q_2S_3=q_1S_2q_3=S_1q_2q_3$} Cross law. These terms give the square of the area of the parallelogram whose sides have quadrances {$q_1$} and {$q_2$} separated by spread {$S_3$}.
• {$S_1S_2q_3=S_1q_2S_3=q_1S_2S_3$} Cross dual law.
• {$S_1S_2S_3$} Triple Spread law.

The squares of the six trigonometric functions find expression as the six possible cross-ratios: q = sin2, 1-q = cos2, 1/q = csc2, 1/(1-q) = sec2, q/(1-q) = tan2, (1-q)/q = cot2. Similarly, the six hyperbolic functions express the quadrance.

Spread law. {$\frac{S_1}{q_1}=\frac{S_2}{q_2}=\frac{S_3}{q_3}$}

Triple quad formula. If {$a_1, a_2, a_3$} collinear, then {$(q_1+q_2+q_3)^2=2(q_1^2+q_2^2+q_3^2)+4q_1q_2q_3$}, which is {$p_1^2=2p_2+4e_3$}.

Triple spread formula. If {$A_1, A_2, A_3$} concurrent, then {$(S_1+S_2+S_3)^2=2(S_1^2+S_2^2+S_3^2)+4S_1S_2S_3$}, which is {$p_1^2=2p_2+4e_3$}.

These formula can be rewritten: {$q_1q_2q_3=(2(q_1^2+q_2^2+q_3^2)-(q_1+q_2+q_3)^2)/4$}, in other words, {$e_3=(2p_2-p_1^2)/4$}

Nonsymmetric formulas

Pythagoras's theorem. If {$a_1a_3\perp a_2a_3$} then {$q_3=q_1+q_2-q_1q_2$}. More elegantly: {$(1-q_3)=(1-q_1)(1-q_2)$}.

Pythagoras's dual theorem. If {$A_1A_3\perp A_2A_3$} then {$S_3=S_1+S_2-S_1S_2$}. More elegantly: {$(1-S_3)=(1-S_1)(1-S_2)$}. Alternatively, {$|\textrm{cos}\,\theta_3|=|\textrm{cos}\,\theta_1||\textrm{cos}\,\theta_2|$}.

{$(q_1q_2S_3-(q_1+q_2+q_3)+2)^2=4(1-q_1)(1-q_2)(1-q_3)$}

Cross law. {$(q_1q_2S_3-(q_1+q_2+q_3)+2)^2=4(1-q_1)(1-q_2)(1-q_3)$}

{$q_1q_2S_3=q_1S_2q_3=S_1q_2q_3= q_1+q_2+q_3 -2 \pm 2\sqrt{(1-q_1)(1-q_2)(1-q_3)}$}

Cross dual law. {$(S_1S_2q_3-(S_1+S_2+S_3)+2)^2=4(1-S_1)(1-S_2)(1-S_3)$}

{$S_1S_2q_3=S_1q_2S_3=q_1S_2S_3= S_1+S_2+S_3 -2 \pm 2\sqrt{(1-S_1)(1-S_2)(1-S_3)}$}

Note that {$\sqrt{1-S_i}=|\textrm{cos}\,\theta_i|$}

Sources

N J Wildberger. Divine Proportions: Rational Trigonometry to Universal Geometry