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(A) : The set of points in one side equidistant from a given line forms a line

Postulate V : If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough

Prove that postulate V implies (A)

Proof : From postulate I through V, we have Euclid's proposition I.34 i.e. in a parallelogram, the opposites and angles are congruent.

How can we finish the proof ?

Add : This is theorem 3.4 in the book geometry from a differential viewpoint - McCleary

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HK Lee
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    Probably (A) has some clarification about the points being on the same side of the given line. As written that set is really two lines. You take two points $A,B$ at the same distance from the given line $\ell$. Call $C,D$ the foots of their perpendiculars to $\ell$. From the congruence of the triangles $ACD$ and $BDC$, and of $ADB$ and $BCA$ conclude that the angle $CAB$ (and $DBA$) are right angles too. Now take a third $E$ point at the same distance and on the same side of $\ell$. The analogous argument gives that angle $EAC$ is also a right angle. – conditionalMethod Oct 23 '19 at 14:08
  • Thank you for your answer. – HK Lee Oct 23 '19 at 14:14

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