Question

write a proof of the theorem that states that through a point not on a line, there is one and only one line parallel to the given line
given: line ( l ) with point ( p ) not on ( l )
prove: there is only one line through ( p ) parallel to ( l )

since ( m parallel l ), ( mangle dcp = \boxed{mangle bpc} ) by the alternate interior angles theorem.

Explanation:

Step1: Recall Parallel Postulate

In Euclidean geometry, the Parallel Postulate (Playfair's Axiom) states: Given a line \( l \) and a point \( P \) not on \( l \), there is exactly one line through \( P \) parallel to \( l \). To prove uniqueness, assume two lines \( m \) and \( n \) through \( P \) are both parallel to \( l \).

Step2: Use Transversal & Angle Properties

Let \( PC \) be a transversal intersecting \( l \) at \( C \) and \( m, n \) at \( P \). By Alternate Interior Angles Theorem, if \( m \parallel l \), \( \angle DCP = \angle BPC \); if \( n \parallel l \), \( \angle DCP = \angle APC \). Thus, \( \angle BPC = \angle APC \), implying \( m \) and \( n \) coincide (since they share \( P \) and an angle, so same line).

Answer:

Through a point \( P \) not on line \( l \), only one line parallel to \( l \) exists (proven via Playfair's Axiom and angle/transversal properties, showing two purported parallels must coincide).