Section 2.1 Triangles

Take the figure above. Because angles \(\alpha\) and \(\gamma\) form a linear pair, the supplement postulate indicates that \(m(\angle\alpha)+m(\angle\gamma)=180\) °. We also know that \(\gamma\) and \(\beta\) form a linear pair, so we know that \(m(\angle\gamma)+m(\angle\beta)=180\) °.

Thus, \(m(\angle\alpha)+m(\angle\gamma)=m(\angle\gamma)+m(\angle\beta)\) So, \(m(\angle\alpha)=m(\angle\beta)\text{.}\) Thus, \(\angle\alpha\cong\angle\beta\text{.}\) We can show by similar processes that \(\angle\gamma\cong\angle\sigma\text{,}\) so the theorem is proven.

Assume there is some arbitrary triangle \(\triangle ABC\) with a point D such that A-B-D, with a line \(\ell\) that passes through point D. We apply the plane separation postulate using \(\ell\) as our line and the fact that a plane contains at least three non-collinear points. By the plane separation postulate, we can create two sets {A,B} and {C}. By the plane separation postulate, this means that segment \(\overline{AC}\) and \(\overline{BC}\) intersect the line \(\ell\text{.}\) Therefore, \(\ell\) must intersect \(\overline{AC}\) or \(\overline{BC}\text{.}\)

Consider \(\triangle ABC\) with \(AB \cong AC\text{.}\)

Draw an angle bisector from \(A\) to the base \(BC\text{,}\) meeting \(BC\) at point \(D\text{.}\)

\(\angle BAD \cong \angle CAD\) from the construction of \(AD\text{.}\)

\(AD \cong AD\) by the reflexive property.

\(\triangle ABD \cong \triangle ACD\) by the Side-Angle-Side congruence postulate.

Corresponding angles of congruent triangles are congruent, so \(\angle B \cong \angle C\text{.}\) Therefore, the angles opposite the congruent sides of an isosceles triangle are congruent.

( \(\Rightarrow\) ) Let \(\overline{BC}\) be a line segment. Construct a perpendicular bisector that intersects \(\overline{BC}\) at a point \(D\) so that \(B-D-C\text{.}\) Place a point \(A\) on this perpendicular bisector. By Axiom 1, we can construct unique line segments \(\overline{AB}\) and \(\overline{AC}\text{.}\) Since \(\overleftrightarrow{AD}\) is a perpendicular bisector to \(\overline{BC}\text{,}\) we have \(\overline{BD}\cong\overline{DC}\text{.}\) By reflexivity, \(\overline{AD}\cong\overline{AD}\text{.}\) In fact, since \(\overleftrightarrow{AD}\) is a perpendicular bisector of \(\overline{BC}\text{,}\) we know that \(m\angle ADB=m\angle ADC=\pi/2\text{.}\) Hence, \(\angle ADB\cong\angle ADC\text{.}\) By the SAS Postulate, \(\triangle ADB\) is in a one-to-one correspondence to \(\triangle ACD\text{.}\) So, \(\overline{AB}\cong\overline{AC}\) and therefore \(|AB|=|AC|\text{,}\) as desired.

Suppose we have \(\triangle ABC\text{.}\) We can construct a ray from \(\angle ACB\) such that it goes through \(\overline{AB}\) and intersects it at point \(E\text{,}\) such that \(A-E-B\) and \(E\) is the midpoint of \(\overline{AB}\text{.}\) Thus, \(\overline{AE} \cong \overline{EB}\text{.}\)

By the Ruler Postulate, we then construct point \(F\) on the ray such that \(C-E-F\text{,}\) with \(E\) as the midpoint of \(\overline{CF}\text{.}\) Therefore, \(\overline{EF} \cong \overline{CE}\text{.}\)

Suppose we have two triangles, \(\triangle ABC\) and \(\triangle DEF\text{,}\) with \(\angle BAC \cong \angle EDF\text{,}\) \(\angle ACB \cong \angle DFE\text{,}\) and \(\overline{AB} \cong \overline{DE}\text{.}\) Assume, for contradiction, that \(m\angle ABC \gt m\angle DEF\text{.}\) We want to construct an angle \(\alpha\) such that \(\angle \alpha \cong \angle E\text{.}\) Using Axiom 18, we can construct a crossbar from \(\angle B\) that intersects \(\overline{AC}\text{,}\) constructing a point \(G\) such that \(A-G-C\text{.}\)

We now also have \(\angle \alpha \cong \angle E\text{.}\) By doing this, through angle-side-angle, the \(\triangle ABG\) that we constructed is congruent to \(\triangle DEF\) and we have \(\angle AGB \cong \angle GCB \cong \angle DFE\text{.}\) But \(\angle AGB\) is an exterior angle of \(\angle BGC\text{,}\) by the exterior angle theorem, \(m\angle AGB \gt m\angle GCB\) and \(m\angle DFE\) which contradicts the definition of congruent triangles.

Essentially, we place \(\triangle DEF\) on \(\triangle ABC\) such that \(D\) overlaps \(A\) and \(E\) overlaps \(B\text{.}\) By axiom 1, we can construct the segment \(\overline{CF}\text{.}\)

We observe the new triangles \(\triangle CDF\) and \(\triangle CBF\text{.}\) Since \(\overline{AC}\cong\overline{DF}\) and \(\overline{BC}\cong\overline{EF}\text{,}\) triangles \(\triangle CAF\) and \(\triangle CBF\) are isosceles. By the Isosceles Triangle Theorem, \(\angle ACF\cong\angle DFC\) and \(\angle BCF\cong\angle EFC\text{.}\)

Take the above figure and suppose that line segments \(i\) and \(j\) to be congruent and the angles \(\angle CBA\) and \(\angle DBA\) to be congruent. Note that line segment \(f\) is congruent to itself. Then, per the Isosceles Triangle Theorem, \(\angle ACB\cong\angle ADB\text{.}\) Thus, by our congruences, we know that \(\triangle ABC\cong\triangle ABD\) per Angle Angle Side congruence.

Case 1: Where 1 side is longer than the others.

Case 2: 2 or more sides are equal

This means for some \(\triangle ABC\) where \(|AB|=|AC|=|BC|\) then for the sum of any two sides, \(2|AB| \gt |AB|\) which proves the theorem.