In reading about triangles, and more especially, in reading derivations or proofs that involve triangles, there are a few words that arise which may not be totally familiar to someone who doesn't read about mathematics daily.
Congruent: The term congruent is used to mean that two geometry objects have exactly the same size and shape. However, they are allowed to have different rotations.
Similar: As the word is applied to triangles and other geometrical shapes, it means that the two objects have the same shape but possibly different sizes. That is to say that one of the objects could be scaled up or down and also rotated so that it exactly matches the other object. For triangles, we can accept these rules:
If two of the three internal angles are equal, then the triangles are similar. (The angle-angle postulate.)
If all three sides are equal, then the triangles are similar. (The side-side-side postulate.)
If two sides and their included angle are equal, then the triangles are similar. (The side-angle-side postulate.)
There are arguments that can be arranged in such a way as to claim a proof for each of these rules, but they are nearly circular in nature and we will simply accept these rules as stated.
The important property about similar triangles is that their angles are all equal and their sides are all proportional.
Concurrent: A set of lines or curves is said to be concurrent if they all meet at a single point. This term is used to define both the centroid and circumcenter of a triangle.
Triangle Area Formula
Although this seems a little basic, there is a nuance that is sometimes overlooked.
$$Area=\frac{1}{2}base\cdot height$$
Figure 1: Area of a Triangle. The figure shows two of the three choices for base and height of the triangle. Note that in the right side graphic, the base is the distance from $A$ to $C$.
The nuance is how to define the base and the height. If one stays inside the triangle, then the base is whichever leg we choose and the height is the perpendicular distance from that base to the opposite vertex. However, one might choose, as the base, a side that makes it impossible to connect the base to the opposite vertex with a perpendicular. In that case, the height is a perpendicular line from the vertex to an extension of the base line far enough away to permit a perpendicular angle.
Heron's Theorem
Heron's theorem is an area formula for any triangle that uses only the side lengths of the triangle. We would hardly ever need it and it is here exclusively for the proof that follows. The proof is somewhat interesting, as it is essentially a derivation.
Heron's Theorem If the sides of any triangle have lengths $a$, $b$, $c$, and we let $s=(1/2)(a+b+c)$ , then $Area=\sqrt{s(s-a)(s-b)(s-c)}$ Proof: For all triangles, there will always be at least one vertex-base combination for which the altitude lies inside the base. For this proof, we will let the base be that side.
From Pythagoras, $$h^{2}+k^{2}=c^{2} \tag{1} \label{1}$$
$$h^{2}+j^{2}=a^{2} \tag{2} \label{2}$$
Using these relationships, the goal is to remove $h,$ $j,$ and $k$ from the equations. From the base length definitions in the triangle, we know that $k=b-j$. Square both sides and expand the result.
$$k^{2}=(b-j)^{2}$$
$$k^{2}=b^{2}-2bj+j^{2}$$
Now add $h^{2}$ to both sides so that we can use $\eqref{1}$ and $\eqref{2}$ to eliminate some terms.
$$h^{2}+k^{2}=b^{2}-2bj+j^{2}+h^{2} \tag{4} \label{4}$$
On the left side of $\eqref{4}$, substitute from $\eqref{1}$ and on the right side, substitute from $\eqref{2}$.
$$c^{2}=b^{2}-2bj+a^{2} \tag{5} \label{5}$$
Next, solve $\eqref{5}$ for j.
$$j=\frac{c^{2}-a^{2}-b^{2}}{-2b}=\frac{a^{2}+b^{2}-c^{2}}{2b} \tag{6} \label{6}$$
Getting from here back to something that looks like Heron's formula is a lot of algebra and there are multiple routes.
From $\eqref{2}$ $h^{2}=a^{2}-j^{2}=(a+j)(a-j)$. Then by substitution from $\eqref{6}$,
$$h^{2}=\left(a+\frac{a^{2}+b^{2}-c^{2}}{2b}\right)\left(a-\frac{a^{2}+b^{2}-c^{2}}{2b}\right)$$
Multiply the $a$ terms by $2b/2b$ so as to combine the denominators.
$$h^{2}=\left(\frac{2ab}{2b}+\frac{a^{2}+b^{2}-c^{2}}{2b}\right)\left(\frac{2ab}{2b}-\frac{a^{2}+b^{2}-c^{2}}{2b}\right)$$
$$h^{2}=\left(\frac{a^{2}+2ab+b^{2}-c^{2}}{2b}\right)\left(\frac{-a^{2}+2ab-b^{2}+c^{2}}{2b}\right)$$
$$h^{2}=\left(\frac{\left(a+b)^{2}-c^{2}\right)}{2b}\right)\left(\frac{c^{2}-(a-b)^{2}}{2b}\right)$$
From pattern matching, $x^{2}-y^{2}=(x-y)(x+y)$ so
$$h^{2}=\left(\frac{(a+b+c)(a+b-c)}{2b}\right)\left(\frac{[c-(a-b)][c+(a-b)]}{2b}\right)$$
$$h^{2}=\left(\frac{(a+b+c)(a+b-c)}{2b}\right)\left(\frac{(-a+b+c)(a-b+c)}{2b}\right)$$
Then multiplying the terms together.
$$h^{2}=\frac{(a+b+c)(a+b-c)(-a+b+c)(a-b+c)}{4b^{2}}$$
In the theorem, we said to let $2s=(a+b+c)$, so we make some clever substitutions.
$$h^{2}=\frac{2s(2s-2c)(2s-2a)(2s-2b)}{4b^{2}}$$
$$h^{2}=\frac{2\cdot2\cdot2\cdot2s(s-a)(s-b)(s-c)}{4b^{2}}=\frac{4s(s-a)(s-b)(s-c)}{b^{2}}$$
$$h=\frac{2}{b}\sqrt{s(s-a)(s-b)(s-c)}$$
Finally, $Area=(1/2)bh$, so
$$Area=\sqrt{s(s-a)(s-b)(s-c)}\qquad\qquad\square$$
30 60 90 Triangle
Since $\pi/2$ is $90^{\circ}$, the three angles of a $30^{\circ}$,$60^{\circ}$,$90^{\circ}$ triangle are
$$\frac{\pi/2}{3}=\frac{\pi}{6}=30^{\circ}$$
$$\frac{2}{3}\cdot\frac{\pi}{2}=\frac{\pi}{3}=60^{\circ}$$
$$\frac{\pi}{2}=90^{\circ}$$
If $h$ is the hypotenuse, then from the definition of sine and a calculator
$$\sin(30^{\circ})=\frac{1}{2}=\frac{opposite}{hypotenuse}$$
Therefore, solving for opposite, the opposite side has length $h/2$.
We can work out the long side from the Pythagorean theorem.
$$h^{2}=\left(\frac{h}{2}\right)^{2}+L^{2}$$
$$L=\frac{h\sqrt{3}}{2}$$
1.8.5 Isosceles Triangles
Figure 2: Isosceles Triangle
When a triangle has two sides of the same length, it is isosceles. Very often, we know one of the two identical sides as well as one angle. That makes solving for the unknown base easy. In figure 2 is shown an isosceles triangle with side lengths of $r$ and base $x$ and one angle, $\alpha.$
Since the angles of a triangle always add to $\pi$, the two angles not given are easily found. Name the two common angles, $\theta$. Using the law of sines, we can write
$$\frac{sin\theta}{r}=\frac{sin\alpha}{x}$$
$$x=\frac{r\cdot sin\alpha}{sin\theta}$$
Once $x$ and $r$ are known, then the Pythagorean relation can be used to obtain the height and then the area.
$$h=\sqrt{r^{2}-(x/2)^{2}}$$
1.8.6 Ceva's Theorem
Figure 3: Ceva's Theorem Triangle. Going around the triangle from $A$ to $B$ to $C$ and back to $A$, the ratio of segments for each side will multiply together to give the value $1$.
Consider an arbitrary triangle with vertices $ABC$ and an arbitrarily chosen point, $P$, somewhere in its interior. Next construct three lines between point $P$ and the three vertices. Each of the lines will intersect the side opposite the vertex from which it is drawn. Label these $DEF$ as in figure 3. Ceva's Theorem asserts that the segment lengths ratio together and that the ratios multiply to yield $1$.
$$\frac{|A-F|}{|F-B|}\cdot\frac{|B-D|}{|D-C|}\cdot\frac{|C-E|}{|E-A|}=1$$
