Unique Parallel Through A Point

Theorem: A unique parallel can be drawn to a given line through a point not lying on the line.

Prerequisites: 
Corresponding angles property (proof)
Transitivity of parallelism (proof)

proof:

Let CD be a given line, and M be a given point. Let us draw a transversal EF passing through the point M and intersecting line CD at a point N such that the angle $\angle END = \theta$.

Now let us draw a line AB through the point N, such that $\angle EMB = \theta$. Since, $\angle EMB = \angle END$, then by the corresponding angles property, AB$\parallel$CD.
Therefore, there exists at least one line passing through the point M, which is parallel to CD.

Now, let us prove uniqueness of the parallel line through M.
For this, if possible, let us assume that A'B' be any other line passing through the point M and parallel to line CD. Since, AB$\parallel$CD and A'B'$\parallel$CD, therefore by transitivity of parallelism, AB$\parallel$A'B'.

But AB intersects A'B' at M, hence there is a contradiction.

Thus, the line AB is unique.


Recommended:
Unique perpendicular theorem
Angle Sum Property of polygon
SAS congruence