Problem
Extend the side $AB$ of an isosceles triangle $ABC$ with $A$ as the vertex, take a point $D$ such that $AB=BD$, and let $E$ be the midpoint of $AB$, then
$$CD=2CE.$$
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Solution
Let the midpoint of the side $AC$ be F.
Then, since $△ABC$ is an isosceles triangle,
$$CE=FB.$$
Since $△ACD$ and $△AFB$ share $∠A$,
$$AD∶AB=2∶1 \qquad and \qquad AC∶AF=2∶1,$$
$$∴ \ △ACD∼△AFB,$$
$$∴ \ CD∶FB=2:1,$$
$$∴ \ CD=2FB,$$
$$∴ \ CD=2CE.$$
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Reference Teiichiro Sasabe (1976) The Encyclopedia of Geometry (2nd edition), Seikyo-Shinsha, p.33.