Problem
If $D$ is the point where the bisector of $∠A$ intersects the side $BC$ in the triangle $ABC$, then
$$AB>BD \qquad and \qquad AC>CD.$$
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Solution
Regarding the triangle $ABD$,
$$∠ADB=∠CAD+∠C,$$
$$∴ \ ∠ADB=∠BAD+∠C, \quad (∵ \ ∠BAD=∠CAD)$$
$$∴ \ ∠ADB>∠BAD,$$
$$∴ \ AB>BD.$$
Similarly, for the triangle $ACD$,
$$∠ADC=∠BAD+∠B,$$
$$∴ \ ∠ADC=∠CAD+∠B,$$
$$∴ \ ∠ADC>∠CAD,$$
$$∴\ AC>CD.$$
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Reference
Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.18