Two conditions must both hold: all corresponding angles are equal, and all corresponding sides are in the same ratio. For triangles specifically, proving the angles match is enough on its own — equal angles automatically force the sides into proportion.

Similarity is a multiplicative relationship, not an additive one — it's a uniform scaling, so every length must be multiplied (or divided) by the same factor. If a $3\times4$ rectangle is enlarged by adding 2 cm to every side, you get $5\times6$ — but $3:4 \ne 5:6$, so the result isn't actually similar to the original.

Always compute it as $k=\dfrac{\text{Big length}}{\text{Small length}}$, so $k>1$ every time — this removes any ambiguity about direction.

• To find a length on the larger shape: multiply the small shape's length by $k$.
• To find a length on the smaller shape: divide the large shape's length by $k$.

Logic check: your answer for a larger-shape length must come out bigger than what you started with, and vice versa — if it doesn't, you've multiplied/divided the wrong way round.

The most reliable method: corresponding sides are always opposite corresponding (matching) angles. If diagrams are rotated or reflected, redraw both shapes separately in the same orientation — this turns a confusing spatial problem into a simple matching exercise.

For nested triangles (a smaller triangle inside a larger one, created by a line parallel to one side), remember that a side of the larger triangle is often the sum of two segments (e.g. $AC=AB+BC$) — don't mistake the visible segment for the whole side.

For "hourglass" triangles (two triangles touching at one vertex between parallel lines), prove similarity first using angle facts: vertically opposite angles at the crossing point, and alternate angles created by the parallel lines.

Worked Example

$AB$ is parallel to $DE$. $ACE$ and $BCD$ are straight lines. $AB=6\text{ cm}$, $AC=8\text{ cm}$, $CD=13.5\text{ cm}$, $DE=9\text{ cm}$. Find $CE$ and $BC$.

1. Prove similarity: $\angle ACB=\angle DCE$ (vertically opposite); $\angle BAC=\angle CED$ and $\angle ABC=\angle CDE$ (alternate angles, since $AB\parallel DE$). So $\triangle ABC \sim \triangle EDC$.
2. Scale factor: using known pair $DE$ and $AB$: $k=\dfrac{DE}{AB}=\dfrac{9}{6}=1.5$.
3. Find $CE$ (on the larger triangle $EDC$): $CE = AC\times k = 8\times1.5=12\text{ cm}$.
4. Find $BC$ (on the smaller triangle, a reverse problem): $BC = CD \div k = 13.5\div1.5=9\text{ cm}$.