An IBL Introduction to Geometries

We prove the contrapositive. Suppose that lines \(AB\) and \(CD\) intersect at a point \(X\) and that points \(E\) and \(F\) lie on lines \(AB\) and \(CD\) respectively so that \(A-E-B\) and \(C-F-D\text{.}\) Then line \(EF\) is a transversal of lines \(AB\) and \(CD\) such that \(\angle EFC\) is an exterior angle of \(\triangle EFX\text{.}\) Then, by the Exterior Angle Theorem, \(m\angle EFC > m\angle FEX\text{.}\) But these two angles are alternate interior angles of the transversal formed by line \(EF\text{,}\) hence if two lines intersect, then the alternate interior angles formed by their transversal are not equal.

Consider a point \(A\) and a line containing \(B\text{,}\) where \(B\) is at the foot of the perpendicular. We want to show \(|AB|\) is the shortest distance, so we need to prove that for any point \(C\) along the line, that \(|AB| \lt |AC|\text{.}\) Construct a triangle with these points using the Distance and Angle postulates so they have the same known distance, where \(\overline{AB}\) bisects \(\overleftrightarrow{CB}\text{.}\) To show \(|AB| \lt |AC|\text{,}\) let's show that \(|AB| \neq |AC|\) and that \(|AB|\) is not greater than \(|AC|\text{.}\)

Case 1: Suppose to the contrary that \(|AB| = |AC|\text{.}\) This would imply (by Theorem 2.1.8) that \(\angle ACB \cong \angle ABC\text{.}\) However, we know \(\angle ABC = 90^\circ\) (by our bisection), so \(\angle ACB = 90^\circ\text{,}\) too. Extend \(\overline{BC}\) (opposite the side of the bisection) to include another point, \(D\text{.}\) By SMSG15, we know \(\angle ACD = 90^\circ\text{,}\) but Theorem 2.1.11 states the exterior angle must be greater than the opposing ones, so \(\angle ACD > \angle ABD\text{,}\) but \(90^\circ\) is not greater than \(90^\circ\text{,}\) so our original assumption was false. So \(|AB| \neq |AC|\text{.}\)

Case 2: Suppose to the contrary that \(|AB| > |AC|\text{.}\) Similarly, we know \(\angle ABC = 90^\circ\) by our bisection. By Theorem 2.1.19, the larger side (compared to \(|AC|\)) \(|AB|\) is opposite the larger angle; that is, \(\angle ACB > \angle ABC\text{.}\) Extend \(\overline{AC}\) (opposite side of the bisection) to include a point \(D\text{.}\) \(\angle ABC = 90^\circ\text{,}\) so \(\angle ACB > 90^\circ\text{,}\) which implies \(\angle BCD \lt 90^\circ\) by SMSG15. However, this cannot be the case since Theorem 2.1.11 says the exterior angle \(\angle BCD\) should be greater than \(\angle ABC\) (or \(\angle BAC\)); so, our original assumption was wrong. Thus \(|AB|\) is not greater than \(|AC|\text{.}\) Therefore, \(|AB| \lt |AC|\text{,}\) so the shortest distance to the line is from the point to the foot of the perpendicular.

\(m\angle 3 + m\angle 6 = \pi \) is given. Similarly, \(m\angle 4 + m\angle 5 = \pi \) is given.

\(\angle 3 \) and \(\angle 4 \) are supplementary.

\(\angle 5 \) and \(\angle 6 \) are supplementary.

\(\angle 4 \) and \(\angle 6 \) are alternate interior angles, and \(\angle 3 \) and \(\angle 5 \) are also alternate interior angles, and by theorem 3.1.2, lines 1 and 2 are parallel.

Let \(\boldsymbol T\) be the statement, “The sum of the angles on one side of a transversal of two lines is equal to the sum of two right angles”. Let \(\boldsymbol P\) be the statement, “The lines are parallel.”

In Theorem 3.1.4 we proved \(\boldsymbol T \implies \boldsymbol P\text{.}\) On the other hand, Euclid's postulate states (in relevant part) \(\neg \boldsymbol T \implies \neg \boldsymbol P\text{.}\) Therefore, together they give \(\boldsymbol T \iff \boldsymbol P\text{.}\)

Suppose we're given two parallel lines \(L\) and \(M\) and a transversal intersecting points \(A\) and \(B\text{.}\) Now, construct a perpendicular from \(A\) to a point \(C\text{.}\) Applying Playfair, using the point \(B\text{,}\) construct a line segment \(\overline{BD}\) that is parallel to \(\overline{AC}\text{.}\) Because \(L\) is parallel to \(M\text{,}\) \(\overline{BD}\) also creates a bisection, so \(\angle ACB \cong \angle ADB\text{.}\) Note we have two triangles \(\triangle ABD\) and \(\triangle ABC\text{.}\) Because \(L\) is parallel to \(M\text{,}\) \(\angle DAB \cong \angle ABC\text{.}\) Of course, \(\overline{AB} \cong \overline{AB}\text{,}\) so the two triangles \(\triangle ADB\) and \(\triangle ABC\) are congruent. Therefore, by CPCTC, \(\overline{BD} \cong \overline{AC}\text{,}\) so the distance from \(L\) to \(M\) is always the same.

By contraposition, Euclid's postulate is equivalent to: “If lines \(l\) and \(m\) are parallel, then the angles on one side of a transversal of \(l\) and \(m\) sum to two right angles.” We therefore show that the equidistance of parallel lines implies that the angles on one side of a transversal of parallels sum to two right angles.

Let \(l\) and \(m\) be parallel lines. By the Equidistance Postulate, if \(X\) is any point on \(l\text{,}\) then \(\mathrm{d(X,m)} = k\) is a constant corresponding to the length of a line segment drawn from \(X\) to the foot of the perpendicular on \(m\text{.}\) Let \(t\) be a transversal of \(l\) and \(m\text{.}\) These line segments form right angles with both \(l\) and \(m\text{.}\)

Let \(t\) intersect \(l\) and \(m\) at \(A\) and \(C\) respectively. Let \(D\) and \(B\) be the points on \(l\) and \(m\) such that \(\overline{AB}\) and \(\overline{CD}\) have length \(k\text{.}\) Then the triangles \(\triangle ABC\) and \(\triangle CDA\) are congruent right triangles by Angle-Angle-Side.

Therefore, if lines \(l\) and \(m\) are parallel, then angles on one side of a transversal of \(l\) and \(m\) sum to two right angles.