Zoom in on a photo and everything grows by the same amount, so nothing looks stretched. Shrink a blueprint and the building keeps its shape. That is similarity: same shape, different size. In this chapter you will learn how to enlarge and shrink figures with dilations, how to prove that two triangles are similar, and how to use similar triangles to find lengths and areas that are hard to measure directly.
1. Dilations and scale factor
A dilation with center \(O\) and scale factor \(k>0\) sends every point \(P\) to the point \(P'\) on ray \(OP\) such that \(OP' = k\cdot OP\). The center \(O\) does not move.
If \(k>1\) the dilation is an enlargement. If \(0
In the coordinate plane, a dilation centered at the origin sends \((x,y)\) to \((kx,ky)\). For a center \((a,b)\) that is not the origin, the image of \((x,y)\) is \((a+k(x-a),\; b+k(y-b))\).
Triangle \(PQR\) has \(P(1,1)\), \(Q(3,1)\), \(R(2,2)\). Dilate it from the origin with \(k=2\).
\(P'=(2,2)\), \(Q'=(6,2)\), \(R'=(4,4)\). Side \(PQ\) has length 2 and \(P'Q'\) has length 4, so lengths doubled, exactly as the scale factor says. Both segments are horizontal, so they are parallel.
2. Similar figures and proportions
Two figures are similar (written \(\sim\)) if one can be mapped onto the other by a dilation followed by a sequence of rigid motions (translations, reflections, rotations). Equivalently, all pairs of corresponding angles are congruent and all pairs of corresponding sides are proportional. The common ratio of corresponding side lengths is the scale factor \(k\).
The order of the letters matters. Writing \(\triangle ABC \sim \triangle DEF\) tells you that \(A\) matches \(D\), \(B\) matches \(E\) and \(C\) matches \(F\), so
\[\dfrac{DE}{AB}=\dfrac{EF}{BC}=\dfrac{DF}{AC}=k.\]
For triangles, one condition is enough (you will see why in Section 3). For figures with more sides you need both: equal angles and proportional sides. A 2 by 6 rectangle and a 4 by 8 rectangle have four right angles each, but \(\tfrac{4}{2}\neq\tfrac{8}{6}\), so they are not similar.
- Match the corresponding sides using the order of the letters.
- Write a proportion with one unknown.
- Cross-multiply and solve, then check that the answer is reasonable (bigger figure, bigger side).
\(\triangle ABC \sim \triangle DEF\) with \(AB=6\text{ cm}\), \(BC=10\text{ cm}\) and \(DE=9\text{ cm}\). Find \(EF\).
The scale factor is \(k=\dfrac{DE}{AB}=\dfrac{9}{6}=1.5\). So \(EF = 1.5\times 10 = 15\) cm.
3. AA, SSS, and SAS similarity
- AA: if two angles of one triangle are congruent to two angles of another, the triangles are similar.
- SSS: if the three pairs of corresponding sides are proportional, the triangles are similar.
- SAS: if two pairs of corresponding sides are proportional and the angles between them are congruent, the triangles are similar.
AA works because the angle sum of a triangle is \(180^\circ\): once two angles match, the third must match as well. Notice that, unlike congruence, you never need to know that a side is equal, only that sides are in the same ratio.
Triangle 1 has sides 6, 9, 12 and triangle 2 has sides 8, 12, 16. Ratios: \(\tfrac{8}{6}=\tfrac{12}{9}=\tfrac{16}{12}=\tfrac{4}{3}\). All equal, so the triangles are similar by SSS with \(k=\tfrac43\).
Now compare sides 4, 6, 9 with 6, 9, 12. The ratios are \(1.5,\ 1.5\) and \(\tfrac{12}{9}\approx 1.33\). They are not all equal, so these triangles are not similar.
When you test SSS, always pair the smallest side with the smallest, the middle with the middle and the largest with the largest. Mixing them up is the fastest way to a wrong answer on my home planet and on yours.
4. The triangle proportionality theorem
If a line parallel to one side of a triangle meets the other two sides at two different points, it divides those sides proportionally. In triangle \(ABC\) with \(D\) on \(AB\), \(E\) on \(AC\) and \(DE\parallel BC\):
\[\dfrac{AD}{DB}=\dfrac{AE}{EC}\qquad\text{and}\qquad \dfrac{AD}{AB}=\dfrac{AE}{AC}=\dfrac{DE}{BC}.\]
The converse is also true: if \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\), then \(DE\parallel BC\).
Why does it work? Angle \(A\) is shared, and \(\angle ADE\cong\angle ABC\) because they are corresponding angles for parallel lines. By AA, \(\triangle ADE\sim\triangle ABC\), and the proportions follow.
In the figure, \(AD=4\), \(DB=6\), \(AE=6\) and \(DE=5\). Find \(EC\) and \(BC\).
\(\dfrac{4}{6}=\dfrac{6}{EC}\), so \(EC=\dfrac{6\cdot 6}{4}=9\). Then \(AB=10\), and \(\dfrac{AD}{AB}=\dfrac{DE}{BC}\) gives \(\dfrac{4}{10}=\dfrac{5}{BC}\), so \(BC=12.5\).
\(\dfrac{AD}{DB}\) compares two pieces of one side. \(\dfrac{AD}{AB}\) compares a piece with the whole side. Never mix them in one proportion: \(\dfrac{AD}{DB}\) pairs with \(\dfrac{AE}{EC}\), while \(\dfrac{AD}{AB}\) pairs with \(\dfrac{DE}{BC}\).
5. The angle bisector theorem
If \(AD\) bisects angle \(A\) of triangle \(ABC\) with \(D\) on \(BC\), then
\[\dfrac{BD}{DC}=\dfrac{AB}{AC}.\]
The bisector splits the opposite side in the same ratio as the two sides that form the angle.
In triangle \(ABC\), \(AB=9\), \(AC=15\), \(BC=16\), and \(AD\) bisects \(\angle A\). Find \(BD\) and \(DC\).
Let \(BD=x\), so \(DC=16-x\). Then \(\dfrac{x}{16-x}=\dfrac{9}{15}\), so \(15x=9(16-x)\), \(24x=144\) and \(x=6\). Hence \(BD=6\) and \(DC=10\). Check: \(6+10=16\) and \(\tfrac{6}{10}=\tfrac{9}{15}\).
6. Writing similarity proofs
- Mark the given information on a sketch.
- Look for hidden angle facts: a shared angle, vertical angles, right angles, and alternate or corresponding angles formed by parallel lines.
- Choose a criterion (AA, SSS or SAS) and list the facts that support it, each with a reason.
- State the conclusion with the letters in matching order.
- If you need a length, write the proportion of corresponding sides and solve.
Given: segments \(AB\) and \(CD\) cross at \(E\), and \(AC\parallel BD\). Prove \(\triangle EAC\sim\triangle EBD\).
| Statement | Reason |
|---|---|
| \(\angle CAE\cong\angle DBE\) | Alternate interior angles, since \(AC\parallel BD\) |
| \(\angle AEC\cong\angle BED\) | Vertical angles are congruent |
| \(\triangle EAC\sim\triangle EBD\) | AA similarity (statements 1 and 2) |
Because the triangles are similar, \(\dfrac{EA}{EB}=\dfrac{EC}{ED}=\dfrac{AC}{BD}\).
7. Scale factor, perimeter, and area ratios
If two figures are similar with scale factor \(k\), then
the ratio of their perimeters (and of any corresponding lengths such as heights or medians) is \(k\);
the ratio of their areas is \(k^2\).
For a triangle, the area formula \(\tfrac12 bh\) multiplies a base by \(k\) and a height by \(k\), so the area is multiplied by \(k\cdot k=k^2\).
| Scale factor \(k\) | Ratio of perimeters | Ratio of areas |
|---|---|---|
| 2 | 2 | 4 |
| 3 | 3 | 9 |
| \(\tfrac12\) | \(\tfrac12\) | \(\tfrac14\) |
| 2.5 | 2.5 | 6.25 |
A 5-12-13 triangle (perimeter 30 in., area 30 in2) is dilated with \(k=1.5\). The new perimeter is \(30\times1.5=45\) in. and the new area is \(30\times1.5^2=67.5\) in2. Check: the new sides are 7.5, 18, 19.5 and \(\tfrac12\cdot 7.5\cdot 18=67.5\).
If a scale factor is 3, areas are multiplied by 9, not by 3. In a model built at a scale of 1 : 8, lengths are 8 times smaller than the real ones, so areas are 64 times smaller.
Key takeaways
- A dilation with scale factor \(k\) multiplies lengths by \(k\) and keeps angle measures; from the origin, \((x,y)\mapsto(kx,ky)\).
- Similar figures have congruent corresponding angles and proportional corresponding sides.
- Triangles are similar by AA, SSS or SAS.
- If \(DE\parallel BC\), then \(\tfrac{AD}{DB}=\tfrac{AE}{EC}\) and \(\tfrac{AD}{AB}=\tfrac{DE}{BC}\); the converse holds too.
- An angle bisector gives \(\tfrac{BD}{DC}=\tfrac{AB}{AC}\).
- Perimeters scale by \(k\), areas by \(k^2\).
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