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Note that $\tan 15^{\circ }=2-\sqrt{3}=\frac{EC}{10}\Rightarrow EC=20-10\sqrt{3}$. (It is important to memorize the sin, cos, and tan values of $15^{\circ }$ and $75^{\circ }$.) Therefore, we have $DE=10\sqrt{3}$. Since $ADE$ is a $30-60-90$ triangle, $AE=2\cdot AD=2\cdot 10=\boxed{20}$