Two figures are similar ($\sim$) if a sequence of rigid motions and dilations carries one onto the other. This builds on Congruence and Rigid Motions (rigid motions only) by also allowing a change of size, using the dilations of Properties of Dilations.
Why it works
Rigid motions keep lengths and angles. A dilation keeps angles and multiplies every length by the same $k$. So a similarity transformation keeps every angle and multiplies every length by one scale factor. For similar triangles $\triangle ABC \sim \triangle DEF$ that means: corresponding angles are congruent, and $\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA} = k$.
Key idea
Similar figures have congruent corresponding angles and proportional corresponding sides. The ratio of corresponding lengths is the scale factor, and it applies to every length, including perimeters.
Worked example: Finding a missing side
$\triangle ABC \sim \triangle DEF$ with $AB = 6$, $BC = 9$ and $DE = 10$. Find $EF$.
- The scale factor from $ABC$ to $DEF$ is $\frac{DE}{AB} = \frac{10}{6} = \frac{5}{3}$.
- $EF = \frac{5}{3} \cdot 9 = 15$.
Answer: $EF = 15$
Worked example: Describing the transformation
$\triangle ABC$ has $A(1, 1)$, $B(3, 1)$, $C(1, 2)$. $\triangle DEF$ has $D(-1, 4)$, $E(5, 4)$, $F(-1, 7)$. Describe a similarity transformation from $ABC$ to $DEF$.
- Compare sides: $AB = 2$ and $DE = 6$; $AC = 1$ and $DF = 3$. Scale factor 3, and the shapes face the same way.
- Dilate by 3 about the origin: $A \to (3, 3)$, $B \to (9, 3)$, $C \to (3, 6)$.
- Each image is 4 left and 1 up from its target... check: $(3, 3) \to (-1, 4)$ is $\langle -4, 1 \rangle$. Translate by $\langle -4, 1 \rangle$: $B \to (5, 4)$, $C \to (-1, 7)$.
Answer: Dilate by 3 about the origin, then translate by $\langle -4, 1 \rangle$.
Worked example: Are all rectangles similar?
Are a $2 \times 3$ rectangle and a $4 \times 9$ rectangle similar?
- All angles are $90^\circ$, so only the sides matter. Short sides: $\frac{4}{2} = 2$. Long sides: $\frac{9}{3} = 3$.
- The ratios differ, so no single scale factor works.
Answer: No.
Common mistake
Matching sides out of order. In $\triangle ABC \sim \triangle DEF$, side $\overline{AB}$ matches $\overline{DE}$ because of the letter order, just as with congruence. If you compare $AB$ with $EF$, you get a wrong scale factor. Check: in a scale factor larger than 1, every side of the bigger figure should be paired with a smaller side of the other.