A triangle with two equal sides is called isosceles; the remaining side is called the base.
Assume \(\triangle ABC\) is an isosceles triangle with the base \([AB]\). Then
\(\measuredangle ABC \equiv - \measuredangle BAC.\)
Moreover, the converse holds if \(\triangle ABC\) is nondegenerate.
The following proof is due to Pappus of Alexandria.
\(CA = CB\), \(CB = CA\), \(\measuredangle ACB \equiv -\measuredangle BCA\).
By Axiom IV,
\(\triangle CAB \cong \triangle CBA.\)
\(\measuredangle BAC \equiv -\measuredangle ABC.\)
To prove the converse, we assume that \(\measuredangle CAB \equiv - \measuredangle CBA\). By ASA condition (Theorem 4.2.1), \(\triangle CAB \cong \triangle CBA\). Therefore, \(CA = CB\).
A triangle with three equal sides is called equilateral.
Let \(\triangle ABC\) be an equilateral triangle. Show that
\(\measuredangle ABC = \measuredangle BCA = \measuredangle CAB.\)
Apply Theorem 4.3.1 twice