Problem
The point $P$ on the bisector of $∠BAC$ is equidistant from its two sides $AB$ and $AC$.
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Solution
Drop a perpendicular line from the point $P$ to $AB$ and $AC$, and let the feet be $E$ and $F$ respectively.
For two right triangles $⊿APE$ and $⊿APF$, they share the side $AP$, and
$$∠EAP=∠FAP,$$
$$∠PEA=∠PFA (=∠R),$$
$$∴ \ ⊿APE≡⊿APF.$$
$$∴ \ PE=PF.$$
That is, the point $P$ is equidistant from $AB$ and $AC$.
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Reference
Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.6