Also, because ABD is a straight line, [Hypothesis.
the angle DBE is equal to the angle EBA.
Therefore the angle DBE is equal to the angle CBE, [Ax. 1.
the less to the greater; which is impossible. [Axiom 9.
Wherefore two straight lines cannot have a common segment.
PROPOSITION 12. PROBLEM.
To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.
Let AB be the given straight line, which may be produced to any length both ways, and let C be the given point without it: it is required to draw from the point C a straight line perpendicular to AB.
Take any point D on the other side of AB, and from the centre C, at the distance CD, describe the circle EGF, meeting AB at F and G. [Postulate 3.
Bisect FG at H, [I. 10.
and join CH.
The straight line CH drawn from the given point C shall be perpendicular to the given straight line AB.
Join CF, CG.
Because FH is equal to HG, [Construction.
and HC is common to the two triangles FHC, GHC;
the two sides FH,HC are equal to the two sides GH, HC, each to each;
and the base CF is equal to the base CG; [Definition 15.
therefore the angle CHF is equal to the angle CHG; [I. 8.
and they are adjacent angles.
But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle, and the straight line which stands on the other is called a perpendicular to it. [Def. 10.
Wherefore a perpendicular CH has been drawn to the given straight line AB from the given point C without it. q.e.f.
