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.

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

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

Note that the order of vertices is important. It is possible that ${\displaystyle \triangle ABC}$ and ${\displaystyle \triangle ACB}$ 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.

Let's say we have two triangles, ${\displaystyle \triangle ABC}$ and ${\displaystyle \triangle DEF}$, 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:

${\displaystyle {\begin{aligned}\angle F+\angle E+\angle D&=180^{\circ }{\text{ (}}\angle {\text{ sum of }}\triangle {\text{)}}\\90^{\circ }+60^{\circ }+\angle D&=180^{\circ }\\\angle D&=180^{\circ }-90^{\circ }-60^{\circ }\\&=30^{\circ }\end{aligned}}}$

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