Triangles similarity test

Two triangles are similar if they have exactly the same shape but not necessarily the same size.
That means all three pairs of angles are equal and the sides are in the same ratio.

The quickest test is AA (Angle-Angle).
If two angles of one triangle match two angles of the other, the triangles are similar.
You only need two, because the third angle is then forced (all three add to $180^\circ$).

Order matters when you name similar triangles.
Writing $\triangle ABC \sim \triangle DEF$ means $A$ matches $D$, $B$ matches $E$, and $C$ matches $F$.
Then the matching sides line up: $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}$.

This proportion is how you find missing lengths.
If one triangle is a scaled copy of another, set up the ratio of corresponding sides and solve.
Corresponding sides are the ones opposite equal angles.

One classic case: in a right triangle, drop an altitude from the right angle to the hypotenuse.
This creates two smaller triangles, each similar to the original and to each other.
The altitude equals the square root of the two pieces it splits the hypotenuse into.

Triangle $ABC$ has angles $50^\circ$ and $60^\circ$. Triangle $DEF$ also has angles $50^\circ$ and $60^\circ$. Are they similar?
Two pairs of angles are equal, which is the AA test.
So yes, $\triangle ABC \sim \triangle DEF$.

Given $\triangle ABC \sim \triangle DEF$ with $AB = 6$, $DE = 3$, and $BC = 8$, find $EF$.
Matching sides are in proportion: $\frac{AB}{DE} = \frac{BC}{EF}$.
So $\frac{6}{3} = \frac{8}{EF}$, which gives $EF = 4$.

In a right triangle, the altitude to the hypotenuse splits it into pieces of length $4$ and $9$. How long is the altitude?
The altitude is the square root of the product of the two pieces.
So it is $\sqrt{4 \times 9} = \sqrt{36} = 6$.