1.4 Justifications and First Theorems
In justifying steps in proofs we will use definitions, postulates, previously proved theorems, and algebraic properties. Here are some algebraic properties:
Properties of Equality
Addition
Property |
If $a=b$
and $c=d$, then $a+c=b+d$ |
(Add. Prop.) |
Mulitplication
Property |
If $a=b$ and $c=d$, then $ac=bd$ |
(Mult. Prop.) |
Distributive Property |
$a(b+c)=ab+ac$ |
(Distrib.
Prop.) |
Reflexive Property |
$a=a$ |
(Reflexive) |
Transitive |
If $a=b$ and $b=c$, then $a=c$ |
(Transitive) |
Substitution Property |
If $a=b$, then either may be
substituted for the other. |
(Substitution) |
Abbreviations and short-hand notations are given in the right-hand column.
Since subtraction is the addition of an additive inverse, the addition property also covers subtraction. Since division is multiplication by a reciprocal, the multiplication property also covers division. Some of these properties apply to congruence as well:
Properties of Congruence
Reflexive
Property |
$\overline{AB}\cong\overline{AB}$ |
(Reflexive) |
Transitive
Property |
If
$\overline{AB}\cong\overline{CD}$, and
$\overline{CD}\cong\overline{CD}\cong\overline{EF}$, then
$\overline{AB}\cong\overline{EF}$ |
(Transitive) |
The Reflexive Property is used when a side is a member of two different triangles. So to say that $\overline{AB} \cong \overline{AB}$ is really to say that $\overline{AB}$ of $\triangle ABC$ is congruent to $\overline{AB}$ of $\triangle ABD$.
We'll illustrate the use of these properties in some simple proofs.
- Theorem 1.1: (The Midpoint Theorem): If $M$ is the midpoint of $\overline{AB}$, then $2AM=AB$ and $AM=\dfrac{1}{2}AB$.
Given: $M$ is the midpoint of $\overline{AB}$.
Prove: $2AM=AB$ and $AM=\dfrac{1}{2}AB$
\begin{array}{ l l | l}
\text{Proof:} & \text{Statements} & \text{Reasons} \\
\hline
& 1. M \text{ is the midpoint of } \overline{AB} & 1. \text{ Given} \\
& 2. AM=MB & 2. \text{ Definition of midpoint} \\
& 3. AM+MB=AB & 3. \text{ Segment Addition Postulate}\\
& 4. AM+AM=AB & 4. \text{ Substitution}\\
& 5. 2(AM)=AB & 5. \text{ Distributive Property}\\
& 6. AM=\dfrac{1}{2}AB & 6. \text{ Multiplication Property}\\
\end{array}
Notice that the hypothesis of the if-then statement is the Given, the information that is assumed to be true at the outset. The conclusion is the Prove.
Here is an analogous theorem for angles:
- Theorem 1.2: (The Angle Bisector Theorem) If $\overrightarrow{BD}$ is the bisector of $\angle ABC$, then $2m\angle ABD=m\angle ABC$ and $m\angle ABD=\dfrac{1}{2}m\angle ABC$.
Given: $\overrightarrow{BD}$ is the bisector of $\angle ABC$
Prove: $2m\angle ABD=m\angle ABC$ and $m\angle ABD=\dfrac{1}{2}m\angle ABC$
\begin{array}{ l l | l}
\text{Proof:} & \text{Statements} & \text{Reasons} \\
\hline
& 1. \overrightarrow{BD} \text{ is the bisector of } \angle{ABC} & 1. \text{ Given} \\
& 2. m\angle ABD=m\angle CBD & 2. \text{ Definition of angle bisector} \\
& 3. m\angle ABD+m\angle CBD=m\angle ABC & 3. \text{ Angle Addition Postulate}\\
& 4. m\angle ABD+ m\angle ABD=m\angle ABC & 4. \text{ Substitution}\\
& 5. 2(m\angle ABD)=m\angle ABC & 5. \text{ Distributive Property}\\
& 6. m\angle ABD=\dfrac{1}{2}m\angle ABC & 6. \text{ Multiplication Property}\\
\end{array}
We now add a couple of definitions and then prove one more theorem — one that will be quite useful.
- Definitions: Complementary angles are two angles whose measures sum to 90.
Note the spelling of "complementary." "Complimentary" angles are always congratulating each other. The problem set will have lots of numerical problems involving complementary and supplementary angles. To solve them, represent one of the angles by a variable, say $x$, express the other angles in terms of $x$, and then look for the equality.
Example: The measure of the complement of an angle is 9 less than twice the measure of the angle. Find the measure of the angle.
Solution: Let $x$ be the measure of the angle, making $90-x$ equal to the measure of the complement. Then
\begin{align*}
90-x&=2x-9\\
3x&=99\\
x&=33
\end{align*}
Example: The measure of the supplement of an angle is three times the measure of the complement. Find the measure of the angle.
Solution: Let $x$ be the measure of the angle, making $180-x$ equal to the measure of the supplement.
\begin{align*}
180-x&=3(90-x)\\
180-x&=270-3x\\
2x&=90\\
x&=45
\end{align*}
- Definitions: Vertical angles are two angles whose sides form pairs of opposite rays.
- Theorem 1.3: (The Vertical Angle Theorem) Vertical angles are congruent.
Given: $\angle 1$ and $\angle 2$ are vertical angles.
Prove: $m\angle 1=m\angle 2$
\begin{array}{ l l | l}
\text{Proof:} & \text{Statements} & \text{Reasons} \\
\hline
& 1. m\angle 1+m\angle 3=180,\;m\angle 2+m\angle 3=180 & 1. \text{ Angle Addition Postulate} \\
& 2. m\angle 1+m\angle 3=m\angle 2 +m\angle 3 & 2. \text{ Substitution} \\
& 3. m\angle 1=m\angle 2 & 3. \text{ Addition Property of Equality}
\end{array}
Note: In a similar fashion we could have shown that $m\angle 3=m\angle 4$. Remember that since we define $\angle 1\cong\angle 2$ to mean $m\angle 1=m\angle 2$ and vice-versa, then proving the one is equivalent to proving the other. However, we cannot write $\angle 1 +\angle 2=180$ since by $\angle 1$we mean a set of points, not a number.
The Vertical Angle Theorem will turn out to be quite useful both in proofs and in numerical problems.
Example: Let $x$ and $y$ be positive integers. Find all ordered pairs $(x,y)$ such that $m\angle 1=10x-30$ and $m\angle 2=50-20y$.
Solution: By the Vertical Angle Theorem, $m\angle 1=m\angle 2$, so
\begin{align*}
10x-30&=50-20y\\
10x+20y&=80\\
x+2y&=8
\end{align*}
Pick values for $y$ to find $x$ giving $(2,3)$, $(4,2)$, and $(6,1)$. But remember that both $10x-30$ and $50-20y$ must be postive, so reject $(2,3)$.