Checking all six parts is a lot of work. For triangles, two pairs of congruent angles are enough. The proof uses the definition of similarity from Similarity Through Transformations and the angle sum from Triangle Sum and Exterior Angle Proofs.
Why it works
Suppose $\angle A \cong \angle D$ and $\angle B \cong \angle E$. Dilate $\triangle ABC$ by $k = \frac{DE}{AB}$. Its image $\triangle A'B'C'$ has $A'B' = DE$, and the dilation keeps angles, so $\angle A' \cong \angle D$ and $\angle B' \cong \angle E$. Now $\triangle A'B'C' \cong \triangle DEF$ by ASA, so a rigid motion carries it onto $\triangle DEF$. The dilation followed by that rigid motion is a similarity transformation: $\triangle ABC \sim \triangle DEF$.
Key idea
AA Similarity: if two angles of one triangle are congruent to two angles of another, the triangles are similar. (The third angles then match automatically.)
Worked example: A line parallel to one side
In $\triangle ABC$, $\overline{DE} \parallel \overline{BC}$ with $D$ on $\overline{AB}$ and $E$ on $\overline{AC}$. $AD = 4$, $DB = 6$ and $DE = 5$. Find $BC$.
- $\angle A$ is shared, and $\angle ADE \cong \angle ABC$ (corresponding angles). So $\triangle ADE \sim \triangle ABC$ by AA.
- Match whole side to whole side: $AB = 4 + 6 = 10$, so the scale factor from $ADE$ to $ABC$ is $\frac{10}{4} = \frac{5}{2}$.
- $BC = \frac{5}{2} \cdot 5 = 12.5$.
Answer: $BC = 12.5$
Worked example: Shadows
A $6$ ft person casts a $4$ ft shadow at the same moment a tree casts a $30$ ft shadow. How tall is the tree?
- The sun's rays hit both at the same angle, and both stand straight up, so the two right triangles are similar by AA.
- $\frac{\text{height}}{\text{shadow}}$ is the same: $\frac{6}{4} = \frac{h}{30}$, so $h = \frac{6 \cdot 30}{4} = 45$.
Answer: $45$ ft
Worked example: An hourglass
$\overline{AE}$ and $\overline{BD}$ cross at $C$, with $\overline{AB} \parallel \overline{DE}$. $AC = 6$, $CE = 9$ and $AB = 8$. Find $DE$.
- $\angle ACB \cong \angle ECD$ (vertical angles) and $\angle A \cong \angle E$ (alternate interior angles). So $\triangle ACB \sim \triangle ECD$ by AA.
- Scale factor from the small triangle to the large: $\frac{CE}{AC} = \frac{9}{6} = \frac{3}{2}$.
- $DE = \frac{3}{2} \cdot 8 = 12$.
Answer: $DE = 12$
Common mistake
Using a part of a side where the whole side belongs. In the first example, $\frac{BC}{DE}$ must equal $\frac{AB}{AD}$ (whole side over whole side), not $\frac{DB}{AD}$. Check: write the triangles' names in matching order, $\triangle ADE \sim \triangle ABC$, and read the sides from them.