3.2 Contents

• Universal Properties
• Types & Terminology
• Intersections
• Vivaldi's Theorem
• Right Angle (Pythagorean) Theorem
• Heron's Formula
• Distance Formula
• Right Angle Definitions

3.2.1 Universal Properties

Closed planar objects with three straight sides, and a sum of lengths of any two sides greater than the third

Perimeter	$a+b+c$
Area (non-obtuse)	$b·h/2$
Sum of Angles	$\pi$

3.2.2 Types & Terminology

Right (all others are oblique)
Obtuse
Acute
Equilateral
Isosceles
Scalene
Congruent triangles

3.2.3 Intersections

Centroid (G)

Median intersection point

Line segments extend from vertices (corners) to the sides at the midpoint
Orthocenter (H)

Altitude intersection point

Line segments extend from vertices to the sides at which right angles are formed
Circumcenter (O)

Perpendicular bisector intersection point

Line segments extend at right angles from the midpoint of each side
Incenter (I)

Angle bisector intersection point

Line segments extend from vertices dividing the angles of the vertices in half
Nine-Point Circle (N=center)

Circle passing though the midpoint of each side, the point of altitude on each side, and the midpoint of lines between each vertex and the orthocenter
Euler Line

Line passing through G, H, N, and O of all non-equilateral triangles (In equilateral triangles, G, H, N, O and I all overlap in a single point)

3.2.4 Vivaldi's Theorem

In an equilateral triangle, the sum of distances from any inner point extending orthogonally to the sides is equal to the height of the triangle

Proof

Given side length $L$ and height $h$, the triangle area equals the sum of the inner triangle areas with altitudes $a$, $b$, and $c$ $$\frac{L·h}{2}=\frac{L·a}{2}+\frac{L·b}{2}+\frac{L·c}{2}$$ Multiply by $2/L$

3.2.5 Right Angle (Pythagorean) Theorem

$$h^2=x^2+y^2$$

Proof

The square $c^2$ has area equal to all the triangle areas subtracted from the largest square $$c·c=(a+b)(a+b)-4·\Big(\frac{1}{2}·a·b\Big)$$ Expand $$c^2=a^2+2·a·b+b^2-2·a·b$$ Cancel like terms

3.2.6 Heron's Formula

$$A=\sqrt{s·(s-a)(s-b)(s-c)}$$ $$s=(a+b+c)/2\text{ (semi-perimeter})$$

Proof

Given a triangle with sides $\{a,b,c\}$, divide it with an orthogonal line $h$ to form two right triangles, dividing line $b$ into lines $x$ and $b-x$ Apply the right angle theorem to determine the relationships and length of $x$ $$x^2=c^2-h^2\therefore x=\sqrt{c^2-h^2}$$ Apply the right angle theorem to determine the relationships of $b-x$ $$(b-x)^2=a^2-h^2$$ Expand $$b^2-2·b·x+x^2-h^2$$ Substitute $x$ and $x^2$ $$b^2-2·b·\sqrt{c^2-h^2}+c^2-h^2=a^2-h^2$$ Simplify $$b^2-2·b·\sqrt{c^2-h^2}+c^2=a^2$$ Isolate $a^2$, $b^2$ and $c^2$ to one side $$b^2+c^2-a^2=2·b·\sqrt{c^2-h^2}$$ Square $$\big(b^2+c^2-a^2\big)^2=4·b^2·(c^2-h^2)$$ Isolate $h^2$ $$h^2=c^2-\frac{\big(b^2+c^2-a^2\big)^2}{4·b^2}$$ Factor the right into a single fraction $$h^2=\frac{4·b^2·c^2-\big(b^2+c^2-a^2\big)}{4·b^2}$$ Factor the numerator as a squared binomial $$h^2=\frac{(2·b·c+b^2+c^2-a^2)(2·b·c-b^2-c^2+a^2)}{4·b^2}$$ Factor the terms with $b$ and $c$ as squared binomials $$h^2=\frac{\big((b+c)^2-a^2\big)\big(a^2-(b-c)^2\big)}{4·b^2}$$ Factor each term in the numerator as squared binomials $$h^2=\frac{(b+c+a)(b+c-a)(a+b-c)(a-b+c)}{4·b^2}$$ For each negative variable within a term, both add and subtract the variable in the term $$h^2=\frac{(b+c+a)(b+c+a-2·a)(a+b+c-2·c)(a+b-2·b+c)}{4·b^2}$$ Take the square root $$h=\frac{\sqrt{(b+c+a)(b+c+a-2·a)(a+b+c-2·c)(a+b-2·b+c)}}{2·b}$$ Multiply by b/2 to obtain the triangle area $$A=\frac{\sqrt{(b+c+a)(b+c+a-2·a)(a+b+c-2·c)(a+b-2·b+c)}}{4}$$ Factor the $4$ into the radicand and expand it into the terms $$A=\sqrt{\bigg(\frac{b+c+a}{2}\bigg)\bigg(\frac{b+c+a}{2}-a\bigg)\bigg(\frac{a+b+c}{2}-c\bigg)\bigg(\frac{a+b+c}{2}-b\bigg) }$$ Substitute $(a+b+c)/2$ for $s$

3.2.7 Distance Formula of a Line

$$|d|=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$$

Proof

Apply the right angle theorem to $(x_1,y_1)$ and $(x_2,y_2)$

3.2.8 Right Angle Definitions

$$\sin(\theta)=\frac{\text{opp}}{\text{hyp}}\qquad\cos(\theta)=\frac{\text{adj}}{\text{hyp}}$$ $$\csc(\theta)=\frac{\text{hyp}}{\text{opp}}\qquad\sec(\theta)=\frac{\text{hyp}}{\text{adj}}$$ $$\tan(\theta)=\frac{\text{opp}}{\text{adj}}\qquad\cot(\theta)=\frac{\text{adj}}{\text{opp}}$$

Simple Trig Angles

Two simple triangles help identify certain values

Function	$$0$$	$$\pi/6$$	$$\pi/4$$	$$\pi/3$$	$$\pi/2$$
$\sin(\theta)$	$$0$$	$$1/2$$	$$1/\sqrt{2}$$	$$\sqrt{3}/2$$	$$1$$
$\cos(\theta)$	$$1$$	$$\sqrt{3}/2$$	$$1/\sqrt{2}$$	$$1/2$$	$$0$$
$\csc(\theta)$	$$∞$$	$$2$$	$$\sqrt{2}$$	$$2/\sqrt{3}$$	$$1$$
$\sec(\theta)$	$$1$$	$$2/\sqrt{3}$$	$$\sqrt{2}$$	$$2$$	$$∞$$
$\tan(\theta)$	$$0$$	$$1/\sqrt{3}$$	$$1$$	$$\sqrt{3}$$	$$∞$$
$\cot(\theta)$	$$∞$$	$$\sqrt{3}$$	$$1$$	$$1/\sqrt{3}$$	$$0$$