Problem
In $△ABC$, when $AC$ is larger than $AB$, $AD$ connecting any $D$ on $BC$ and $A$ is smaller than $AC$.
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Solution
If $AC>AB$, then from the problem $0038$,
$$∠B>∠C.$$
Since $∠ADC$ is an exterior angle of $△ABD$,
$$∠ADC>∠B,$$
$$∴ \ ∠ADC>∠C,$$
$$∴ \ AC>AD.$$
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Reference
Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.12