Problem
The line segment $BD$ joining the point $B$ of the base of an isosceles triangle $ABC$ with vertex $A$ to any point $D$ on $AC$ is greater than the line segment $DE$ joining the point $D$ to any point $E$ on $BC$.
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Solution
$$∠ABC=∠ACB \qquad and \qquad ∠DBE<∠ABC,$$
$$∴ \ ∠DBE<∠DCE \ (=∠ACB).$$
However, since $∠DEB$ is an exterior angle of $△DCE$,
$$∠DCE<∠DEB,$$
$$∴ \ ∠DBE<∠DEB,$$
$$∴ \ BD>DE.$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.33.