Theorem: The two angles opposite to the equal sides of an isosceles triangle are equal.
Prerequisites:
Isosceles Triangle (definition)
SAS rule of congruency (proof)
Proof:
Let $\triangle ABC$ be an isosceles triangle with AB = AC. Let AD be the angle bisector of $\angle BAC$.
In triangles $\triangle ABD$ and $\triangle ACD$,
$\qquad\quad\angle BAD = \angle CAD\qquad\qquad\qquad\text{(AD bisects $\angle BAC$)}\\
\qquad\quad AB = AC\qquad\qquad\qquad\qquad\quad\:\!\text{(given)}\\
\qquad\quad AD = AD\qquad\qquad\qquad\qquad\quad\text{(common)}$
Hence by SAS rule of congruency, $\triangle ABD\cong\triangle ACD$.
Therefore, by CPCTC (corresponding parts of the congruent triangles are congruent), $\angle B = \angle C$.
Q.E.D.
Recommended:
Converse of Isosceles Triangle Theorem
Basic Proportionality Theorem
Angle Bisector Theorem
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