Properties of similar triangles

Similar triangles have the same shape but can differ in size.
Their corresponding angles are equal, and their corresponding sides are proportional.
These two facts are what "similar" means.

To show two triangles are similar, you usually check the angles.
If two pairs of corresponding angles are equal, the triangles are similar — this is the angle-angle (AA) rule, and the third pair is then automatically equal.
A line parallel to one side of a triangle creates equal angles, so it cuts off a triangle similar to the whole.

It is important to match up the vertices that have the same angle.
The similarity statement lists them in that order: if $\angle A = \angle P$, $\angle B = \angle R$, and $\angle C = \angle Q$, you write $\triangle ABC \sim \triangle PRQ$.
Corresponding sides then join matching vertices, so $\frac{AB}{PR} = \frac{BC}{RQ} = \frac{AC}{PQ}$ — not $\frac{AB}{PQ}$, because $B$ matches $R$, not $Q$.

The ratio of the areas is the square of the ratio of the sides.
If corresponding sides are in ratio $\frac{1}{3}$, the areas are in ratio $\left(\frac{1}{3}\right)^2 = \frac{1}{9}$.
So a triangle with sides $3$ times as long has $9$ times the area.

A line parallel to one side gives a ready-made pair of similar triangles.
If $DE$ is parallel to $BC$ in $\triangle ABC$, then $\triangle ADE \sim \triangle ABC$.
Their corresponding sides are proportional, so $\frac{AD}{AB} = \frac{DE}{BC}$.

In $\triangle LMN \sim \triangle XYZ$, angle $N$ is $80^\circ$. What is angle $Z$?
The similarity statement matches $N$ with $Z$, since they are in the same position.
Corresponding angles are equal, so $\angle Z = 80^\circ$.

In $\triangle ABC \sim \triangle PRQ$, side $AB = 6$ and side $PR = 9$. What is the ratio $\frac{BC}{RQ}$?
Because $B$ matches $R$ and $C$ matches $Q$, side $BC$ corresponds to side $RQ$.
All corresponding sides share the same ratio, so $\frac{BC}{RQ} = \frac{AB}{PR} = \frac{6}{9} = \frac{2}{3}$.

Triangles $ABC$ and $PQR$ are similar with $AB = \frac{1}{3}PQ$. If the area of $\triangle ABC$ is $k$, what is the area of $\triangle PQR$?
The side ratio is $\frac{1}{3}$, so the area ratio is $\left(\frac{1}{3}\right)^2 = \frac{1}{9}$.
So $\triangle PQR$ has area $9k$.