Playfair's postulate is logically equivalent to the statement: If a line intersects one of two parallel lines, then it also intersects the other
Playfair's postulate is logically equivalent to the statement: If a line intersects one of two parallel lines, then it also intersects the other
Let l and m be parallel. Let t be a line not parallel to l. Then there is a point that they intersect at. Let Q be any point on t. By playfairs postulate there is a unique line through Q that is parallel to l call it m. Since t is any line not parallel to l and p is any point on t, then p is any point not on l. This allows us to use play fair's postulate to produce a line parallel to l (thus all of them). Therefore given two parallel lines and if a third line intersects one of the parallel lines it intersects the other.
Conversely, Given 2 parallel lines l and m and a transversal that intersects both l and m we will show Playfair's postulate. Let l be the line that is given in Playfair's postulate and let P be any point. By contradiction, suppose that the line m which can be constructed parallel to l and through the point P is not unique. There are m and m' through P parallel to l. Let Q and Q' be a point on m and m' respectively such that Q P Q'. Notice by the 3rd Incidence axiom these are non-collinear points, otherwise Q would be on m' and Q' would be on m and m = m'. Drop a perpendicular from Q and Q' to l. Since P, Q, Q' are non-collinear then the segments from QL is either greater or less than Q'L'. (L and L' is a point on the line l) If QL = Q'L' then P, Q and Q' would be collinear and m = m'.
W.L.O.G. Let QL > Q'L' implies m' os betweem m and l and thus does not have P as a point since m, m' and l are parallel. This contradicts the construction of m' thus m' is the unique line through P parallel to l.
Therefore Euclid's 5th postulate equivalent to Playfair's postulate is logically equivalent to the statement that If a line intersect one of two parallel lines, then is also intersects the other.