Isosceles Triangle Theorem

An isosceles triangle is a triangle that has two equal sides.

The isosceles triangle theorem states the following:

Isosceles Triangle Theorem

In an isosceles triangle, the angles opposite to the equal sides are equal. Conversely, if the base angles of a triangle are equal, then the triangle is isosceles.

Consider isosceles triangle $\triangle ABC$ with $AB=AC,$ and suppose the internal bisector of $\angle BAC$ intersects $BC$ at $D.$ Now consider the triangles $\triangle ABD$ and $\triangle ACD$. We have $AB=AC$, $AD=AD$ and $\angle BAD=\angle CAD$ by construction. Hence, $\triangle ABD\cong\triangle ACD$ by the SAS congruence axiom. So $\angle ABC=\angle ACB$. $_\square$

Note: The converse holds, too. If we were given that $\angle ABC=\angle ACB$, in a similar way we would get $\triangle ABD\cong\triangle ACD$ by the AAS congruence theorem. Thus, $AB=AC$ follows immediately.

This theorem gives an equivalence relation. In order to show that two lengths of a triangle are equal, it suffices to show that their opposite angles are equal. In fact, given any two segments $AB$ and $AC$ in the plane with $A$ as a common endpoint, we have $AB=AC\Longleftrightarrow \angle ABC=\angle ACB$. So in a geometry problem, if we are to show equality of two sides of a triangle, we can start chasing angles!

In $\triangle ABC$ we have $AB=AC$ and $\angle ABC=47^\circ$. Find $\angle BAC$.

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By the isosceles triangle theorem, we have $47^\circ=\angle ABC=\angle ACB$. Since the angles in a triangle sum up to $180^\circ$, we have

$$\angle BAC=180^\circ - \left(\angle ABC+\angle ACB\right)=180^\circ-2\times 47^\circ=86^\circ. \ _\square$$