Analyzing the diagram, it can be seen that $\angle 1$ and $\angle 2$ form a straight angle, so these are supplementary angles. Similarly, $\angle 2$ and $\angle 3$ are also supplementary angles.

Therefore, by the Angle Addition Postulate, the sum of $m\angle 1$ and $m\angle 2$ is $180^{\circ },$ and the sum of $m\angle 2$ and $m\angle 3$ is also $180^{\circ }.$ These facts can be used to express $m\angle 2$ in terms of $m\angle 1$ and in terms of $m\angle 3.$

Angle Addition Postulate	Isolate $m\angle 2$
$m\angle 1+m\angle 2$ $=$ $180^{\circ }$	$m\angle 2$ $=$ $180^{\circ }-m\angle 1$
$m\angle 2+m\angle 3$ $=$ $180^{\circ }$	$m\angle 2$ $=$ $180^{\circ }-m\angle 3$

By the Transitive Property of Equality, the expressions representing $m\angle 2$ can be set equal to each other. Then the equation can be simplified. By the definition of congruent angles, this means that the vertical angles $\angle 1$ and $\angle 3$ are congruent angles. The same process can be used to prove $\angle 2$ and $\angle 4$ congruent.

Two-Column Proof

The previous proof can be summarized in the following two-column table.

Statements	Reasons
${\ell }_1$ and ${\ell }_2$ lines	Given
$\angle 1$ and $\angle 2$ supplementary	Definition of straight angle
$m\angle 1+m\angle 2=180^{\circ }$	Definition of supplementary angles
$m\angle 2=180^{\circ }-m\angle 1$	Subtraction Property of Equality
$\angle 2$ and $\angle 3$ supplementary	Definition of straight angle
$m\angle 2+m\angle 3=180^{\circ }$	Definition of supplementary angles
$m\angle 2=180^{\circ }-m\angle 3$	Subtraction Property of Equality
$180^{\circ }-m\angle 1=180^{\circ }-m\angle 3$	Transitive Property of Equality
$m\angle 1=m\angle 3$	Subtraction and Multiplication Properties of Equality

Consider the points $A,$ $B,$ $C,$ and $D$ on each ray that starts at the point of intersection $E$ of the two lines.

Suppose that points $A$ and $B$ are rotated $180^{\circ }$ about point $E.$ The points $A$ and $B$ are mapped onto the points $A^{\prime }$ and $B^{\prime }$ after the rotation. This means that $\angle AEB$ is mapped onto $\angle A^{\prime }E{B}^{\prime }.$ Since rotations are a rigid motion, $\angle AEB$ and $\angle A^{\prime }E{B}^{\prime }$ are congruent angles. Since the point $A^{\prime }$ lies on $\overrightarrow{EC}$ and point $B^{\prime }$ lies on $\overrightarrow{ED},$ $\angle A^{\prime }E{B}^{\prime }$ is congruent to $\angle CED.$ By applying the Transitive Property of Congruence, it can be confirmed that $\angle AEB$ is congruent to $\angle CED.$