Angles, Triangles & Similarity

Angles: the vocabulary you actually need

An angle is two rays sharing an endpoint (the vertex). We measure it in degrees (full circle = 360°) or, in a few lessons, radians. Here are the four terms that recur forever:

• Right angle: exactly 90°. The box symbol means it.
• Complementary: two angles summing to 90°. If one is $\theta$, its complement is $90° - \theta$.
• Supplementary: two angles summing to 180°. If one is $\theta$, its supplement is $180° - \theta$.
• Vertical angles: when two lines cross, the angles across from each other.

Let's actually prove vertical angles are equal, because the Federation of Boring Textbook Authors just asserts it and moves on — goddamn cowards. Suppose two lines cross forming angles $\alpha, \beta, \alpha', \beta'$ around the vertex. The adjacent pair $\alpha$ and $\beta$ share a line, so they're supplementary:

$\alpha + \beta = 180°.$

The pair $\beta$ and $\alpha'$ are also supplementary (other side of the same line):

$\beta + \alpha' = 180°.$

Subtract: $\alpha = \alpha'$. Done. Vertical angles are equal, and it cost us exactly one subtraction — cheaper than the chalk I wasted writing it. Remember this: geometry theorems are often just algebra in disguise.

Parallel lines and a transversal

Now slice two parallel lines with a third line (the transversal). The transversal creates eight angles total — four at each intersection — but by the vertical-angle theorem, we only need to understand two at one crossing to know them all.

The key claim: alternate interior angles are equal, and corresponding angles are equal. These follow from the definition of "parallel" (the lines never meet because they have the same direction), and they are the raw material for the most important theorem in this lesson.

Specifically: if lines $\ell_1$ and $\ell_2$ are parallel and a transversal crosses them, the four "interior" angles between the parallels satisfy:

$\angle_{\text{alt interior 1}} = \angle_{\text{alt interior 2}}, \qquad \angle_{\text{co-interior 1}} + \angle_{\text{co-interior 2}} = 180°.$

The triangle angle sum — a proof, not a promise

Theorem. The three interior angles of any triangle sum to 180°.

Proof. Take triangle $ABC$. Draw a line through vertex $B$ that is parallel to side $AC$ — call it $\ell$. Now we have three angles sitting along $\ell$ at $B$:

• The angle on the left side of $B$: by alternate interior angles with $\ell \parallel AC$, this equals $\angle A$.
• The angle in the middle: that's just $\angle B$ itself.
• The angle on the right side: by alternate interior angles on the other side, this equals $\angle C$.

The three angles fill exactly a straight line at $B$:

$\angle A + \angle B + \angle C = 180°. \qquad \square$

Every single step is justified. No hand-waving, no "it looks like 180°", no cheating with a damn protractor. If you ever doubt it: the alternative would be a geometry where parallel lines don't behave, and we'd need to move to a different planet. File under: hypotheses matter.

Congruent vs similar: two flavors of "same shape"

Congruent triangles ($\cong$) are identical in every measurement — same angles, same side lengths. Think of it as same shape AND same size.

Similar triangles ($\sim$) share only the angles. They are scaled copies of each other. The sides are proportional: if $\triangle ABC \sim \triangle DEF$, then

$\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD}.$

Every ratio equals the same scale factor $k$. Multiply all sides of the original by $k$ and you land on the copy.

The most useful similarity criterion is AA (Angle-Angle): if two triangles share two angle measures, the third is forced (it's $180°$ minus the other two), so all three angles match, so the triangles are similar.

Why AA suffices: you know two angles, so the third is $180° - \text{sum of first two}$. Both triangles have the same three angles, so they're similar. QED.

Why similarity is the engine of trigonometry

Here's the sentence you must tattoo on your brain — and I do not say that lightly, I have contraband chalk and I will write it on the reactor wall if necessary: because all right triangles with the same acute angle are similar, the ratio of any two sides depends only on that angle — not on the size of the triangle.

Draw a right triangle with acute angle $\theta$. Draw any other right triangle with the same $\theta$. By AA (they both have $90°$ and $\theta$, so the third angle is forced to $90° - \theta$), they are similar. Their sides are proportional with some scale factor $k$. But the ratios — opposite over hypotenuse, adjacent over hypotenuse — are unchanged by $k$. That means those ratios are pure functions of $\theta$.

That's why sin, cos, and tan are possible. They measure properties of an angle, not of any particular triangle. File this away — the next two lessons build on it hard.

The exterior angle theorem (a bonus, because it's beautiful)

An exterior angle of a triangle is the angle formed by extending one side. It's supplementary to the interior angle next to it. So if the interior angles are $A$, $B$, $C$, the exterior angle at $C$ is $180° - C$. But $A + B + C = 180°$, so $180° - C = A + B$.

Exterior angle theorem: an exterior angle of a triangle equals the sum of the two remote interior angles. Brief, elegant, and used constantly in proofs.