How do we show that a pair of triangles are similar to one another?

In practice, one does not need to compare all corresponding angles and sides to demonstrate similarity.

Two triangles with one equal angle are not necessarily similar.

For example, \(\triangle ABC\) is not similar to \(\triangle AB'C'\), even though they share \(\angle A\).

Two triangles with two (and, therefore, all three) angles equal must be similar.

For example, \(\triangle ABC\) is similar to \(\triangle AB'C'\) since \(\angle A\) is shared and \(\angle C = \angle C'\).

Two triangles with one angle equal, are not necessarily similar.

For example, \(\triangle ABC\) is not similar to \(\triangle AB'C'\) even though they share \(\angle A\).

Two triangles with two pairs of corresponding sides in the same proportion and the angle adjacent to both of those sides equal are always similar.

For example, \(\triangle ABC\) is similar to \(\triangle AB'C'\), since \(\angle A\) is shared and  \(\dfrac{AB'}{AB} = \dfrac{AC'}{AC}\).

For the side-angle-side (SAS) rule, we must use the angle at the vertex formed by the sides that are used. We call this angle the contained angle.

For example, consider \( \triangle ABC\) and \(\triangle AB'C' \).

Observe:

• \(\triangle ABC\) and \(\triangle AB'C' \) share \(\angle A\),

• \(\dfrac{AB}{AB'}=\dfrac{BC}{B'C'}\), and
• ​​\(\triangle ABC \sim \triangle AB'C' \).

​​​​​​However, \(B'D = B'C'\) so \(\dfrac{AB}{AB'}=\dfrac{BC}{B'D}\), but\(\triangle ABC\) is not similar to \(\triangle AB'D\).​​​​

You should have found that \(\triangle AB'C'\) is initially free to take any shape.

Two triangles with all corresponding sides in the same proportion will always be similar.

For example, \(\triangle ABC\) is similar to \(\triangle AB'C'\), since \(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}=\dfrac{B'C'}{BC}\).