Problem
In $△ABC$, let $AB>AC$, and take any point $P$ on the median line $AD$. Then,
$$AB-AC>PB-PC.$$
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Solution
If we take a point $E$ on $AB$ so that $AC=AE$,
$$AB-AC=AB-AE=EB.$$
Since $AB>AC$ and $D$ is the midpoint of $BC$,
$$∠BAD<∠CAD,$$
$$∴ \ PE<PC.$$
For $△PEB$,
$$PE+EB>PB,$$
$$∴ \ EB>PB-PE>PB-PC,$$
$$∴ \ AB-AC>PB-PC.$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.27.