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# 4.3: Isoseles Triangles

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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