4.3: Isoseles Triangles

A triangle with two equal sides is called isosceles; the remaining side is called the base.

Theorem $\PageIndex{1}$

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.

Proof

Note that

$CA = CB$, $CB = CA$, $\measuredangle ACB \equiv -\measuredangle BCA$.

By Axiom IV,

$\triangle CAB \cong \triangle CBA.$

Applying the theorem on the signs of angles of triangles (Theorem 3.3.1) And Axiom IV again, we get that

$\measuredangle BAC \equiv -\measuredangle ABC.$

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.

Exercise $\PageIndex{1}$

Let $\triangle ABC$ be an equilateral triangle. Show that

$\measuredangle ABC = \measuredangle BCA = \measuredangle CAB.$

Hint

Apply Theorem 4.3.1 twice