Geometry for Elementary School/Similarity

In this chapter, we will start the discussion of similarity and similarity theorems. We say the two figures are similar if they have the same shape but different size. Similar figures have three things in common: corresponding sides (corr. sides), corresponding angles (corr. ∠s) and corresponding points (corr. points). We will only talk about similar triangles.

Congruent triangles

The triangles $\mathrm{△}ABC$ and $\mathrm{△}DEF$ are similar if and only if all the following conditions hold:

1. The side $\overline{AB}$ is proportional to $\overline{DE}$. (Corresponding sides)
2. The side $\overline{BC}$ is proportional to $\overline{EF}$. (Corresponding sides)
3. The side $\overline{AC}$ is proportional to $\overline{DF}$. (Corresponding sides)
4. The angle $\mathrm{\angle }ABC$ equals $\mathrm{\angle }DEF$. (Corresponding angles)
5. The angle $\mathrm{\angle }BCA$ equals $\mathrm{\angle }EFD$. (Corresponding angles)
6. The angle $\mathrm{\angle }CAB$ equals $\mathrm{\angle }FDE$. (Corresponding angles)

Note that the order of vertices is important. It is possible that $\mathrm{△}ABC$ and $\mathrm{△}ACB$ are not similar even though both refer to the same triangle. Remember that the place where corresponding points are must be the same on both triangles.

Similarity theorems give a set of the fewest conditions that are sufficient in order to show that two triangles are similar. They are 3 sides proportional, AAA and ratio of 2 sides, inc. ∠. We will talk about them later on.

Finding the value of unknowns in triangles whose similarity is given

Let's say we have two triangles, $\mathrm{△}ABC$ and $\mathrm{△}DEF$, and they are congruent. AB=3, BC=5, EF=10, ∠F=90° and ∠E=60°. We need to find DE and ∠A. Here's how:

$\begin{array}{cc}\mathrm{\angle }F+\mathrm{\angle }E+\mathrm{\angle }D& =180^{\circ }\text{ (}\mathrm{\angle }\text{ sum of }\mathrm{△}\text{)}\\ 90^{\circ }+60^{\circ }+\mathrm{\angle }D& =180^{\circ }\\ \mathrm{\angle }D& =180^{\circ }-90^{\circ }-60^{\circ }\\ & =30^{\circ }\end{array}$

$\begin{array}{cc}\because \mathrm{△}ABC& \sim \mathrm{△}DEF\text{ (given)}\\ \therefore \frac{DE}{AB}& =\frac{EF}{BC}\text{ (corr. sides, }\sim \mathrm{△}\text{s)}\\ \frac{DE}{3}& =\frac{10}{5}\\ DE& =\frac{10}{5}\times 3\\ DE& =6\\ \mathrm{\&}\mathrm{\angle }A& =\mathrm{\angle }D\text{ (corr. }\mathrm{\angle }\text{s, }\sim \mathrm{△}\text{s)}\\ \mathrm{\angle }A& =30^{\circ }\end{array}$