Similar polygons are polygons with the same shape, but not necessarily the same size; thus, they are dilations of each other. Think of when the eye doctor dilates your eyes to see inside them better; he makes the pupils the same shape, but just makes them larger.
We can use proportions and scale factors to find missing parts of similar polygons. The symbol ~ denotes similarity. Note also that we can always “turn proportions sideways” and they still work.
For generic polygons, in order to prove similarity, you have to show that all pairs of corresponding angles are the same, and all pairs of corresponding sides are ratios of each other. But with triangles, similarity has shortcuts, as we’ll see below.
Similar Triangles
The triangle similarity shortcuts use the following postulates:
- AA Similarity Postulate: Two triangles are similar if two angles of one triangle are congruent to two angles of the second triangle.
- SSS Similarity Postulate: Two triangles are similar if their three corresponding side lengths are proportional.
- SAS Similarity Postulate. Two triangles are similar if two of their corresponding side lengths are proportional and the sides’ included angle are congruent.
Note that if two triangles are congruent, then technically they are also similar. Note also that if you need to flip a triangle for the sides to be proportional, the triangles are still similar.
Here are some examples of similar triangles:
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The triangles are similar via SSS ~ since $ \displaystyle \frac{{13}}{6.5}=\frac{{11}}{5.5}=\frac{{12}}{6}$. |
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The triangles are similar via SAS ~, since $ \displaystyle \frac{{15}}{9}=\frac{{10}}{6}$, and both triangles have an included angles of $ \displaystyle 90{}^\circ $ (right angle). |
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Triangle $ ABC$ is similar to triangle $ CBD$ are similar via SAS ~, since $ \displaystyle \frac{{10}}{4}=\frac{{25}}{10}$, and triangles have congruent included angles. |
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Triangle $ ABD$ is similar to triangle $ CBD$ are similar via AA~. |
Triangle Proportionality Theorem, or Side-Splitter Theorem:
The Triangle Proportionality Theorem, or Side-Splitter Theorem states that a line parallel to one side of a triangle intersecting the other two sides divides those two sides proportionally.
This creates a smaller triangle within the larger triangle, and the two triangles and it also turns out that the two triangles are similar. Here is this visually:
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Think of just making proportions with the different parts of the sides, but make sure they line up:
If $ \displaystyle \overline{{BC}}\,\,||\,\,\overline{{DE}}$, then $ \displaystyle \frac{{\overline{{AB}}}}{{\overline{{BD}}}}=\frac{{\overline{{AC}}}}{{\overline{{CE}}}}$. (Also, $ \displaystyle \frac{{\overline{{AB}}}}{{\overline{{AC}}}}=\frac{{\overline{{BD}}}}{{\overline{{CE}}}}$, by “turning proportions sideways”.) The converse is also true: if $ \displaystyle \frac{{\overline{{AB}}}}{{\overline{{BD}}}}=\frac{{\overline{{AC}}}}{{\overline{{CE}}}}$, then $ \displaystyle \overline{{BC}}\,\,||\,\,\overline{{DE}}$. Also, since the triangles are similar, it also follows that $ \displaystyle \frac{{\overline{{AB}}}}{{\overline{{AD}}}}=\frac{{\overline{{AC}}}}{{\overline{{AE}}}}$. As a consequence of this theorem, we have this the Triangle Midsegment Theorem: |
Proportional Parts of Parallel Lines
A special case of the Triangular Proportionality Theorem involve at least two parallel lines that are cut by two transversals. Since as the non-parallel transversals are extended, triangles are formed:
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If $ \displaystyle \overline{{AD}}\,\,||\,\,\overline{{BE}}\,\,||\,\,\overline{{CF}}$, then $ \displaystyle \frac{{\overline{{AB}}}}{{\overline{{BC}}}}=\frac{{\overline{{DE}}}}{{\overline{{EF}}}}$ (part-to-part). Also, $ \displaystyle \frac{{\overline{{AB}}}}{{\overline{{DE}}}}=\frac{{\overline{{BC}}}}{{\overline{{EF}}}}=\frac{{\overline{{AC}}}}{{\overline{{DF}}}}$ (part-to-part and part-to-whole) and $ \displaystyle \frac{{\overline{{AB}}}}{{\overline{{AC}}}}=\frac{{\overline{{DE}}}}{{\overline{{DF}}}}$ (part-to-whole), and so on.
It’s common sense; just make sure the proportions match similar segments! |
Triangle Angle Bisector Theorem
The Triangle Angle Bisector Theorem says that any angle bisector in a triangle divides the opposite sides into two segments proportional to the lengths of the other two sides of the triangle. This sounds really confusing, so here is an example. Makes sense!
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If $ \displaystyle \angle ABC\cong \angle CBD$, $ \displaystyle \frac{{\overline{{AC}}}}{{\overline{{AB}}}}=\frac{{\overline{{CD}}}}{{\overline{{BD}}}}$, or $ \displaystyle \frac{{\overline{{AB}}}}{{\overline{{AC}}}}=\frac{{\overline{{BD}}}}{{\overline{{CD}}}}$.
Since I prefer to work from top to bottom, I would set it up this way: $ \displaystyle \frac{{10}}{8}=\frac{{25}}{20}$. Works! The converse is also true. |
Geometric Means in Right Triangles
There are some triangle similarity rules that make use of what we call a Geometric Mean. To get a geometric mean of two numbers, just multiple them and then take the positive square root. So, the geometric mean between $ a$ and $ b$ is $ x$ (positive), where $ \displaystyle \frac{{x}}{a}=\frac{{b}}{x}$. For example, the geometric mean between $ 6$ and $ 10$ is $ \displaystyle \sqrt{{6\times 10}}=\sqrt{{60}}=\sqrt{{4\times 15}}=2\sqrt{{15}}$.
Using the Geometric Mean with Similar Right Triangles
There are two methods that are typically used with finding missing parts of right triangle similar triangles: the “Pulse” or “Heartbeat” method and the “Rabbit” method. Here are drawings showing why we come up with those representations:
| Heartbeat:
Find $ x$: |
$ \displaystyle \frac{{x}}{12}=\frac{{12}}{x+7}$
Start from the bottom left, go up and then down like a heartbeat. See how the altitude acts as a geometric mean? In this case, we would solve to get $ x$: $ \displaystyle \begin{array}{l}\displaystyle \frac{x}{{12}}=\displaystyle \frac{{12}}{{x-7}};\,\,\,x\left( {x-7} \right)=144\\{{x}^{2}}-7x-144=0;\,\,\left( {x-16} \right)\left( {x+9} \right)=0\\x=16\,\,\,\,\,\,\,\text{(throw away }-9\text{)}\end{array}$ |
| Rabbit:
Find $ x$ and $ y$: |
$ \displaystyle \frac{{8}}{x}=\frac{{x}}{14}$ or $ \displaystyle \frac{{14}}{x}=\frac{{x}}{8}$ | $ \displaystyle \frac{{6}}{y}=\frac{{y}}{14}$ or $ \displaystyle \frac{{14}}{y}=\frac{{y}}{6}$ See the head and body of the rabbit? And how the altitude isn’t used at all? I always start on the bottom to a side (either part whole), go up and down a side, and then back down to the bottom, using the opposite I used before (part or whole). Note how you can go up and down either side. To get $ x$ and $ y$: $ \displaystyle \begin{array}{l}{{x}^{2}}=14\times 8=112;\,\,\,x=\sqrt{{112}}=4\sqrt{7}\\{{y}^{2}}=14\times 6=84;\,\,\,y=\sqrt{{84}}=2\sqrt{{21}}\end{array}$ |
Problems and Solutions:
Problem: Prove the following using a two-column proof; this one’s a little tricky:
Given: $ ABCD$ is a trapezoid. 
Prove: $ \displaystyle \frac{{\overline{{AF}}}}{{\overline{{CF}}}}=\frac{{\overline{{BF}}}}{{\overline{{DF}}}}$
Solution: 
Proof:
| Statements | Reasons |
| 1. $ ABCD$ is a trapezoid. | 1. Given |
| 2. $ \displaystyle \overline{{AB}}\,\,||\,\,\overline{{DC}}$ | 2. Definition of a trapezoid: bases are parallel |
| 3. $ \displaystyle \angle DCF\cong \angle BAF$ | 3. $ \displaystyle \text{AIA}\Rightarrow \,\cong \,\angle \,\text{ }\!\!’\!\!\text{ s}$ |
| 4. $ \displaystyle \angle AFB\cong \angle CFD$ | 4. Vertical angles are $ \cong $ |
| 5. $ \displaystyle \vartriangle AFB$ ~ $\vartriangle CFD$ | 5. AA ~ |
| 6. $ \displaystyle \frac{{\overline{{AF}}}}{{\overline{{CF}}}}=\frac{{\overline{{BF}}}}{{\overline{{DF}}}}$ | 6. Similar triangles have proportional side lengths. |
Problem:
Find $ x$:  |
Use the side-splitter theorem; think of rotating the triangle so one of the parallel bases is at the bottom.
Set up proportions with the different parts of the sides, but make sure they line up: $\ \displaystyle \frac{{\text{large bottom}}}{{\text{small bottom}}}=\frac{{\text{large side}}}{{\text{small side}}}: \frac{{x+10}}{{x+1}}=\frac{8}{2}$. Cross-multiply to get $ \displaystyle 2x+20=8x+8;\,\,\,6x=12;\,\,\,x=2$. |
Problem:
Find $ x$:  |
Use the Triangle Midsegment Theorem.
It appears that the parallel lines are midsegments of the triangle, since $ \displaystyle 2x+2=2\left( {x+1} \right)$. Then we know that the sides are congruent: $ \displaystyle x+4=2x-6$, or $ x=10$. Try it; it works! |
Problem:
Find $ x$ and $y$:  |
Us the Triangular Proportionality Theorem.
Since the four lines cut by two transversals are parallel, we know that $ \displaystyle \frac{y}{3}=\frac{{y+3}}{4}$, for example, so $ \displaystyle 4y=3y+9;\,\,\,y=9$. We also know that $ \displaystyle \frac{x}{3}=\frac{{2x-5}}{4}$, so $ \displaystyle 4x=6x-15;\,\,\,x=7.5$. |
Problem:
Find $ x$:  |
Since the base of the small triangle form midpoints of the two sides, this is a midsegment of the larger triangle. Because of the Triangle Midsegment Theorem, the bottom base is twice the length of the midsegment, so $ x=14$.
We also know that the two bases of the triangles are parallel. |
Problem:
Find $ x$:  |
Use the Triangle Angle Bisector Theorem.
$ \displaystyle \frac{x+5}{10}=\frac{{6x-6}}{30}$, so $ \displaystyle 30x+150=60x-60;\,\,\,x=7$ |
Problem:
Find $ x$:  |
This is tricky, since we need to use both the “heartbeat” and “rabbit” techniques, Separate the bottom of the triangle (turned on it’s side) to $ y$ and $ 5-y$, so they add up to $ 5$:  .
Now we can apply the two techniques, and solve a system! Tricky! Heartbeat: $ \displaystyle \frac{{5-y}}{x}=\frac{{x}}{y}$; Rabbit: $ \displaystyle \frac{{5-y}}{4}=\frac{{4}}{5}$ Solve the system, using the equation with only $ y$ ‘s (Rabbit) first and cross-multiplying: $ 25-5y=16; y=\displaystyle \frac{{9}}{5}$ yard. Now, solve for $x$, using $\displaystyle \frac{{9}}{5}$ for $ y$: $ \displaystyle \frac{{5-\displaystyle \frac{9}{5}}}{x}=\frac{x}{{\displaystyle \frac{9}{5}}};\,\,{{x}^{2}}=\left( {5-\displaystyle \frac{9}{5}} \right)\left( {\displaystyle \frac{9}{5}} \right)=\left( {\displaystyle \frac{16}{5}} \right)\left( {\displaystyle \frac{9}{5}} \right)=\displaystyle \frac{{144}}{{25}};\,\,x=\displaystyle \frac{{12}}{5}$ yard (positive only). |
On to Right Triangle Geometry.