Two-Column Proofs: Lines, Angles and Perpendicular Bisectors

Where proofs start

Every definition uses other words, and those words need defining too. The chain has to stop somewhere, so geometry begins with a few undefined terms that are described but not defined:

Undefined term	Described as
point	a location, with no size
line	a straight path extending forever in both directions
plane	a flat surface extending forever

Distance along a line and distance around a circular arc are also taken as starting ideas. Everything else is defined from these.

Term	Precise definition
segment $\overline{AB}$	points $A$ and $B$ and every point of the line between them
ray $\overrightarrow{AB}$	$A$, and every point on the line through $B$ on $B$‘s side of $A$
angle	two rays sharing an endpoint, called the vertex
circle	all points in a plane at a fixed distance from a fixed point, the center
perpendicular lines	lines that meet to form a right angle
parallel lines	lines in the same plane that never meet

Precision is the point. Parallel says in the same plane, because two lines in space can fail to meet without being parallel — think of a highway overpass crossing a road below it.

The shape of a two-column proof

A proof states what is given, what is to be proved, and then a chain of statements, each with its reason.

Kind of reason	Example
given	"$\overline{AB} \cong \overline{CD}$" was stated in the problem
definition	a midpoint divides a segment into two equal parts
postulate	through two points there is exactly one line
theorem	vertical angles are congruent, once proved
property of equality	substitution, reflexive, transitive, addition

The format is a discipline, not decoration. Every statement needs a reason written next to it, so a step that only looks right in the picture has nowhere to hide.

Theorem: vertical angles are congruent

Given: lines $\ell$ and $m$ cross, forming $\angle 1$, $\angle 2$ and $\angle 3$, with $\angle 1$ and $\angle 3$ vertical and $\angle 2$ between them. Prove: $\angle 1 \cong \angle 3$.

Statement	Reason
$\angle 1$ and $\angle 2$ form a linear pair	definition of linear pair
$m\angle 1 + m\angle 2 = 180°$	linear pairs are supplementary
$\angle 2$ and $\angle 3$ form a linear pair	definition of linear pair
$m\angle 2 + m\angle 3 = 180°$	linear pairs are supplementary
$m\angle 1 + m\angle 2 = m\angle 2 + m\angle 3$	substitution
$m\angle 1 = m\angle 3$	subtraction property of equality
$\angle 1 \cong \angle 3$	definition of congruent angles

The notation matters here. $m\angle 1$ is the measure of the angle, a number; $\angle 1 \cong \angle 3$ says the angles are congruent. The last line converts one into the other, and it needs its own reason.

Theorem: alternate interior angles

When a line — a transversal — crosses two lines, it forms eight angles. Pairs in matching positions at the two crossings are corresponding; pairs on opposite sides of the transversal, between the two lines, are alternate interior.

The starting point is accepted without proof:

Corresponding angles postulate. If two parallel lines are cut by a transversal, corresponding angles are congruent.

Given: $\ell \parallel m$, cut by a transversal. $\angle 1$ and $\angle 5$ are corresponding; $\angle 5$ and $\angle 3$ are vertical; so $\angle 1$ and $\angle 3$ are alternate interior. Prove: $\angle 1 \cong \angle 3$.

Statement	Reason
$\ell \parallel m$	given
$\angle 1 \cong \angle 5$	corresponding angles postulate
$\angle 5 \cong \angle 3$	vertical angles are congruent
$\angle 1 \cong \angle 3$	transitive property of congruence

The vertical angles theorem proved above is now a reason. That is how geometry builds: each theorem becomes a tool for the next proof, which is why the order they are proved in matters.

Theorem: the perpendicular bisector

The perpendicular bisector of $\overline{AB}$ is the line through its midpoint $M$ at right angles to it.

Given: $P$ is on the perpendicular bisector of $\overline{AB}$, which meets $\overline{AB}$ at $M$. Prove: $PA = PB$.

Statement	Reason
$M$ is the midpoint of $\overline{AB}$	definition of perpendicular bisector
$AM = BM$	definition of midpoint
$\angle PMA$ and $\angle PMB$ are right angles	definition of perpendicular
$\angle PMA \cong \angle PMB$	all right angles are congruent
$PM = PM$	reflexive property
$\triangle PMA \cong \triangle PMB$	SAS
$PA = PB$	CPCTC

The standard says the perpendicular bisector is exactly the set of such points, so the converse needs proving too.

Given: $PA = PB$ with $P$ not on $\overline{AB}$, and $M$ is the midpoint of $\overline{AB}$. Prove: $\overline{PM} \perp \overline{AB}$.

Statement	Reason
$PA = PB$	given
$AM = BM$	definition of midpoint
$PM = PM$	reflexive property
$\triangle PMA \cong \triangle PMB$	SSS
$\angle PMA \cong \angle PMB$	CPCTC
$m\angle PMA + m\angle PMB = 180°$	linear pairs are supplementary
$\angle PMA$ is a right angle	congruent supplementary angles are right angles
$\overline{PM} \perp \overline{AB}$	definition of perpendicular

Together the two proofs show a point is on the perpendicular bisector if and only if it is equidistant from $A$ and $B$.

Why working backwards finds the proof

A proof is written forwards but found backwards.

For the perpendicular bisector theorem, the goal is $PA = PB$. Equal sides are usually corresponding parts of congruent triangles — so the real goal becomes find two congruent triangles containing $PA$ and $PB$. The picture offers $\triangle PMA$ and $\triangle PMB$. Which criterion fits? Two sides and a right angle between them: SAS.

Each question shrinks the problem until what is left is the given. Then the proof is that chain, written in the opposite order.

Worked examples

Common mistakes

Practice problems