Geometry for Elementary School/Congruence

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

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

1. The side ${\displaystyle {\overline {AB}}}$ equals ${\displaystyle {\overline {DE}}}$. (Corresponding sides)
   
   
   
   
2. The side ${\displaystyle {\overline {BC}}}$ equals ${\displaystyle {\overline {EF}}}$. (Corresponding sides)
   
   
   
   
3. The side ${\displaystyle {\overline {AC}}}$ equals ${\displaystyle {\overline {DF}}}$. (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 congruent even though both refer to the same triangle. Remember that the place where corresponding points are must be the same on both triangles.

Congruence theorems give a set of the fewest conditions that are sufficient in order to show that two triangles are congruent. They are SSS, SAS, ASA, AAS and RHS. 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, ∠F=90° and ∠E=60°. We need to find DE and ∠A. Here's how:

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

${\displaystyle {\begin{aligned}\because \triangle ABC&\cong \triangle DEF{\text{ (given)}}\\\therefore DE&=AB{\text{ (corr. sides, }}\cong \triangle {\text{s)}}\\DE&=3\\\And \angle A&=\angle D{\text{ (corr. }}\angle {\text{s, }}\cong \triangle {\text{s)}}\\\angle A&=30^{\circ }\end{aligned}}}$