An acute isosceles triangle, $ABC$, is inscribed in a circle. Through $B$ and $C$, tangents to the circle are drawn, meeting at point $D$. If $\angle{ABC}=\angle{ACB}=2\angle{D}$ and $x$ is the radian measure of $\angle{A}$, then $x=$
$\text{(A) } \frac{3\pi}{7}\quad \text{(B) } \frac{4\pi}{9}\quad \text{(C) } \frac{5\pi}{11}\quad \text{(D) } \frac{6\pi}{13}\quad \text{(E) } \frac{7\pi}{15}$
We can make two equations (assume angle D is y): $y+2x=180$ and $4y+x=180$. We find that $x=\dfrac{540}{7}$. Now we have to convert this to radians. 360 degrees is $2\pi$ radians, so since we have $\dfrac{540}{7}$ degrees, the answer is $\dfrac{3\pi}{7}$ which is $\fbox{A}.$