Problem
If the bisector of the vertex angle $A$ of a triangle $ABC$ is perpendicular to the base $BC$, then the triangle is an isosceles triangle.
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Solution
If the intersection point of the bisector of $∠A$ and $BC$ is $D$, then $△ABD$ and $△ACD$ share the side $AD$,
$$∠BAD=∠CAD \qquad and \qquad ∠BDA=∠CDA \ (=∠R),$$
$$∴ \ △ABD≡△ACD,$$
$$∴ \ AB=AC.$$
Therefore, $△ABC$ is an isosceles triangle.
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.32.