2.4 Perpendicularity
We start with the following postulate.
- Postulate: The
Perpendicular Postulate
In a plane, through a point on a line, there is one and only one line perpendicular to the given line.
- Definition: In a plane, the perpendicular bisector of a segment is the line that is perpendicular to the segment at its midpoint.
Armed with these two ideas we can now prove an important assertion about perpendicular bisectors.
- Theorem 2.12:
The Perpendicular Bisector Characterization Theorem (PBC)
In a plane, the perpendicular bisector of a segment is the set of all points that are equidistant from the endpoints of the segment.
We break this theorem into two if-then statements.
Given: In a plane,
$k$ is the perpendicular bisector of $\overline{AB}$. Prove: $PA=PB$ |
\begin{array}{ l l | l}
\text{Proof:} & \text{Statements} & \text{Reasons} \\
\hline
& 1. \text{ Line $k$ is the perpendicular bisector of }\overline{AB} & 1. \text{ Given} \\
&2. M \text{ is the midpoint of }\overline{AB} & 2. \text{ Definition of perpendicular bisector}\\
S&3. AM=MB & 3. \text{ Definition of midpoint}\\
S&4. PM=PM & 4. \text{ Reflexive}\\
&5. \angle AMP,\;\angle BMP\text{ are right angles} & 5. \text{ Definition of perpendicular lines}\\
A&6. \angle AMP\cong\angle BMP & 6. \text{ All right angles are congruent}\\
&7. \triangle AMP\cong\triangle BMP & 7. \text{ SAS}\\
&8. PA=PB & 8. \text{ CPCTC}\\
\end{array}
Given: $P$ and
$\overline{AB}$ lie in a plane with $PA=PB$ Prove: $P$ lies on the perpendicular bisector of $\overline{AB}$. |
The proof of this theorem can be most easily expressed in paragraph form since it involves two cases.
Case 1: |
Suppose $P$
lies
on $\overline{AB}$; then since $PA=PB$, $P$ is the midpoint of
$\overline{AB}$ and thus it lies on the perpendicular bisector by
definition. |
Case 2: |
Suppose $P$
lies
off $\overline{AB}$. Let $M$ be the midpoint of
$\overline{AB}$. It is given that $PA=PB$. By the reflexive
property $PM=PM$ and since $M$ is the midpoint of $\overline{AB}$, then
$AM=MB$. Then by SSS, $\triangle PAM\cong \triangle PBM$, making
$\angle PMA\cong\angle PMB$. Since congruent adjacent angles
formed by intersecting lines are right angles,
$\overleftrightarrow{PM}\perp\overline{AB}$, so $P$ lies on the
perpendicular bisector of $\overline{AB}$. |
The next proof is instructive since it shows a number of ways in which auxiliary lines can be used. It is also an existence proof, demonstrating that a certain line exists.
- Theorem 2.13:
If a point is not on a line, then there is a line through the point
perpendicular to the given line.
Given: Point $P$ does
not lie on line $\overleftrightarrow{AB}$. Prove: There is a line through $P$ perpendicular to $\overleftrightarrow{AB}$. |
\begin{array}{ l l | l}
\text{Proof:} & \text{Statements} & \text{Reasons} \\
\hline
& 1. \text{ Point $P$ does not lie on line }\overleftrightarrow{AB}. & 1. \text{ Given} \\
&2. \text{ Draw auxilary line }\overleftrightarrow{AP} & 2. \text{ Two points determine a line.}\\
A&3. \text{ Draw $\overrightarrow{AC}$ so that } m\angle CAB=m\angle PAB & 3. \text{ Protractor Postulate}\\
S&4. \text{ Let $D$ be a point on $\overrightarrow{AC}$ such that }AD=AP. & 4. \text{ Ruler Postulate}\\
&5. \text{ Draw auxiliary line }\overleftrightarrow{PD} & 5. \text{ Two points determine a line.}\\
S& 6. \overline{AE}\cong\overline{AE} &6. \text{ Reflexive}\\
&7. \triangle DAE\cong \triangle PAE &7. \text{ SAS}\\
&8.\angle PEA\cong \angle DEA &8. \text{ CPCTC}\\
&9. \overline{PD}\perp\overleftrightarrow{AB} &9. \text{ If two lines form congruent adjacent angles,}\\
&& \quad\text{then the lines are perpendicular.}\\
\end{array}
A final crucial theorem is the following.
- Theorem 2.14:
In a plane, if each of two points is equidistant from the endpoints of
a segment, then the line they lie on is the perpendicular bisector of
the segment.
We will ask you to prove this in the exercise set, but you may use it in the other problems.