An IBL Introduction to Geometries

The line \(\overleftrightarrow{AB}\) intersects the line \(\overleftrightarrow{CD}\) at point E. \(\angle AEC\) and \(\angle BEC\) are supplementary. \(\angle BEC\) and \(\angle BED\) are also supplementary. \(180^{\circ} - \angle BEC = \angle AEC = \angle BED\text{.}\) So, \(\angle AEC \cong \angle BED\text{.}\) Thus verticle angles are congruent.

Let line \(l\) intersect \(\triangle ABC\) at point \(D\) so \(A - D - B\text{.}\)

Let line \(l = \overline{DE}\text{.}\) There are many ways for point \(E\) to be placed.

Case 1: \(E\) lies outside of \(\triangle ABC\text{.}\) If we can form two convex sets \({D, B}\) and \({E}\) then given line \(\overline{AC}\text{,}\) we can use the plane separation postulate to find that \(\overline{DE}\) intersects \(\overline{AC}\) since \(D\) and \(E\) are in different sets.

Case 2: \(E\) again lies outside of \(\triangle ABC\text{.}\) If we can form two convex sets \({D, A}\) and \({E}\) then given line \(\overline{BC}\text{,}\) we can use the plane separation postulate to find that \(\overline{DE}\) intersects \(\overline{BC}\) since \(D\) and \(E\) are in different sets.

Case 3: \(E\) lies within \(\triangle ABC\text{.}\) The Ruler Postulate states that the distance between two points is the absolute value of the difference of the corresponding real numbers. If \(l\) were to curve, then the distance between two points may be less than the difference of their real numbers, therefore \(l\) is a straight line. Since \(l\) cannot curve, \(l\) must intersect \(\overline{AC}\) or \(\overline{BC}\text{.}\)

Case 4: \(E\) lies outside \(\triangle ABC\text{.}\) If we can form two convex sets \({E}\) and \({C}\) and we are given line \(\overline{AB}\text{,}\) then since line \(l\) must enter the triangle \(\triangle ABC\text{,}\) we have covered this case in Case 3 and line \(l\) must intersect \(\overline{AC}\) or \(\overline{BC}\text{.}\)

In all cases, line \(l\) intersects \(\overline{AC}\) or \(\overline{BC}\text{.}\)

Let \(\triangle ABC\) be an isoceles triangle where, without loss of generality, \(\overline{AB} \cong \overline{BC}\text{.}\) Then, since \(m\angle ABC = m\angle CBA\text{,}\) so \(\angle ABC \cong \angle CBA\text{.}\) So we have \(\overline{AB} \cong \overline{BC}\) and \(\angle ABC \cong \angle CBA\text{,}\) so by the SAS Postulate, there is a congruence relation between corresponding parts of \(\triangle ABC\) and \(\triangle CBA\text{.}\) Therefore, \(\angle BAC \cong \angle BCA\text{.}\) So the angles opposite the equal sides are congruent.

Consider the points \(A, B, C,\) and \(D\) such that \(\overline{BC}\) forms a segment, and \(\overline{AD}\) forms an arbitrary segment bisecting \(\overline{BC}\text{.}\) We want to show \(D\) is equidistant from the endpoints \(B\) and \(C\text{;}\) that is, \(\overline{AB} \cong \overline{AC}\text{.}\) First, note for the triangle created, because \(\overline{AD}\) bisects \(\overline{BC}\text{,}\) \(\overline{BD} \cong \overline{DC}\text{,}\) and \(m\angle BDA = m\angle CDA\text{.}\) Also, \(\overline{AD} \cong \overline{AD}\text{,}\) so a Side-Angle-Side correspondence can be made. This implies the triangles \(\triangle ABD\) and \(\triangle ADC\) are congruent, which means (by definition) \(\overline{AB} \cong \overline{AC}\) (This—of course—can be applied no matter where A lies, without loss of generality). Conversely, suppose a point is equidistant from the endpoints; that is, \(\overline{AB} \cong \overline{AC}\text{.}\) Now, because \(\overline{AB} \cong \overline{AC}\text{,}\) an isosceles (potentially equilateral) triangle is formed, and by the Isosceles Triangle Theorem, the angles opposite the sides are equivalent, too. If we bisect \(\overline{BC}\) such that \(B-D-C\text{,}\) we also know that \(\overline{BD} \cong \overline{BC}\text{.}\) As previously stated, we know that \(m\angle ABD = m\angle ACD\text{.}\) Once again, we've found a Side-Angle-Side correspondence, implying \(\Delta ABD \cong \Delta ACD\text{.}\) Since these two triangles are congruent, we know that \(m\angle BDA = m\angle CDA\text{,}\) so \(A\) lies on the bisector segment.

Let \(\triangle ABC\) be a triangle, and let \(D\) be a point on \(\overline{AC}\) such that \(A-C-D\text{.}\) Then, \(\angle BCD\) is an exterior angle of \(\triangle ABC\) by the definition of an exterior angle. We will also construct a point \(E\) on \(\overline{BC}\) using the Ruler Postulate such that \(\overline{BE} \cong \overline{EC}\text{.}\) We can use the Ruler Postulate again to place a point \(F\) on \(\overline{AE}\) such that \(\overline{AE} \cong \overline{EF}\text{.}\) We can construct a one-to-one correspondence between the triangles \(\triangle AEB\) and \(\triangle FEC\text{.}\) Since \(\overline{BE} \cong \overline{EC}\text{,}\) \(\overline{AE} \cong \overline{EF}\text{,}\) and \(\angle AEB \cong \angle FEC\) (by vertical angle congruence), we know that by the SAS Postulate, the correspondence is a congruence correspondence. So, \(\angle ECF \cong \angle EBA\text{.}\) And, \(\angle EBA \cong \angle CBA\) by the Angle Construction Postulate. Now, \(F\) is on the interior of \(\angle BCD\text{,}\) so by the Angle Addition Postulate, \(m\angle BCF < m\angle BCD\text{,}\) and thus \(m\angle CBA < m\angle BCD\text{.}\)

A similar construction can be done to show that, for a point \(D^\prime\) constructed such that \(B-C-D^\prime\text{,}\) \(m\angle CAB < m\angle ACD^\prime\text{.}\)

Suppose we have two triangles \(\triangle ABC\) and \(\triangle DEF\) such that \(\overline{AC} \cong \overline{DF}\text{,}\) \(\angle BAC \cong \angle EDF\text{,}\) and \(\angle BCA \cong \angle EFD\text{.}\) We will ultimately want to prove this using Side-Angle-Side correspondence, so let's show \(\overline{AB} \cong \overline{DE}\text{.}\) Trivially, if \(\overline{AB} \cong \overline{DE}\) already, we're done, and a Side-Angle-Side correspondence can be found - hence the two triangles are congruent. However, if \(\overline{AB} \ncong \overline{DE}\text{,}\) without loss of generality, consider the case where \(\overline{DE} > \overline{AB}\text{.}\) Using the Ruler Postulate, move the point \(X\) (lying on the line segment \(\overline{DE}\) such that \(D-X-E\)) along \(\overline{DE}\) until \(\overline{DX} \cong \overline{AB}\text{.}\) Note that \(\angle BCA \cong \angle XFD\) because we have a Side-Angle-Side correspondence. Note that \(\overline{DX} \cong \overline{AB}\text{,}\) \(\angle BAC \cong \angle EDF\text{,}\) and \(\overline{AC} \cong \overline{DF}\text{.}\) By CPCTC, because the two triangles \(\triangle ABC\) and \(\triangle DXF\) are congruent, then \(\angle BCA \cong \angle XFD\text{.}\) But we were given that \(\angle BCA \cong \angle EFD\text{,}\) so \(\angle XFD \cong \angle EFD\) - using the Angle Construction Postulate. This means we never needed to shift the point \(X\text{,}\) so \(\overline{AB} \cong \overline{DE}\text{.}\) And becuase \(\overline{AC} \cong \overline{DF}\) and \(\angle BAC \cong \angle EDF\text{,}\) we have a Side-Angle-Side correspondence - thus the two triangles \(\triangle ABC\) and \(\triangle DEF\) are congruent. Therefore, if - for two triangles - we know two corresponding angles and the side between them are congruent, then both triangles are congruent. Conversely, if two triangles are congruent, by definition we know that two corresponding angles and a side between them are congruent. Thus, the biconditional result is shown.

Suppose we have one triangle \(\triangle ABC\) such that \(\angle ABC \cong \angle ACB\text{.}\) Construct an angle bisecting segment \(\overline{AD}\) (so \(\angle BAD \cong \angle CAD\)); this can be done by Crossbar Theorem, and Angle Measurement Postulate should tell us \(\angle ABC \cong \angle ACB\text{.}\) Now, we know that \(\overline{AD} \cong \overline{AD}\) (by the Ruler and Distance Postulates). Because \(\angle BAD \cong \angle CAD\text{,}\) \(\angle ABD \cong \angle ADB\) (Angle Construction Postulate), and \(\overline{AD} \cong \overline{AD}\text{,}\) we have an Angle-Angle-Side correspondence between the two triangles \(\triangle ABD\) and \(\triangle ACD\text{.}\) Therefore, \(\triangle ABD \cong \triangle ACD\text{,}\) and by CPCTC, \(\overline{AB} \cong \overline{AC}\text{.}\) So, we've proven the result that the two sides opposite the congruent angles in an isosceles triangle are congruent.

The third case requires additional argument, since we cannot use the same set of inequalities to show that \(|AB|+|AC| > |BC|\text{.}\) Instead, we construct an additional triangle. First, extend line \(AB\text{.}\) Then mark a point \(D\) at a distance \(|AC|\) from point \(A\) in the direction opposite \(B\) (by the Ruler Postulate). Then \(|BD| = |AB| + |BC|\text{.}\)

Draw line segment \(CD\text{.}\) Then in \(\triangle BCD\text{,}\) \(\angle ADC\) is opposite side \(\overline{BC}\) and \(\angle BCD\) is opposite side \(\overline{BD}\text{.}\) We show that \(m\angle BCD > m\angle ADC\text{.}\) By the previous theorem, this implies that \(|BD| = |AB| + |BC| > |BC|\text{.}\)