Problem
Take any point $D$ on the side $AB$ of $△ABC$, and any point $F$ on the extension of $AC$.
Connecting $D$ and $F$, let $N$ be the intersection of the bisectors of $∠ADF$ and $∠ABC$, and let $M$ be the intersection of the bisectors of $∠AFD$ and $∠ACB$. Then,
$$∠BND=∠CMF.$$
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Solution
Since $N$ is an excenter of $△DPB$, from the problem $0077$,
$$∠BND=\frac{1}{2}∠DPB.$$
Similarly, since $M$ is an excenter of $△FCP$,
$$∠CMF=\frac{1}{2}∠FCP.$$
Since $∠DPB=∠FCP$,
$$∠BND=∠CMF.$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.21