$ 解：(1)设\angle BAC={n}^{\circ }。$
$ 由题意得\pi \cdot DE=\dfrac {n\pi \cdot AD}{180}，AD=2DE，$
$ \therefore n=90，即\angle BAC={90}^{\circ }。$
$ (2)\because AD=2DE=10(\ \mathrm {cm})，$
$ \therefore BC=20\ \mathrm {cm}。$
$ \therefore {S}_{阴}=\dfrac{1}{2}BC\cdot AD-{S}_{扇形AEF}$
$ =\dfrac {1}{2}\times 10\times 20-\dfrac {90\pi \cdot {10}^2}{360}$
$ =(100-25\pi ){\ \mathrm {cm}}^2。$
