Right triangle $ACD$ with right angle at $C$ is constructed outwards on the hypotenuse $\overline{AC}$ of isosceles right triangle $ABC$ with leg length $1$, as shown, so that the two triangles have equal perimeters. What is $\sin(2\angle BAD)$? $\textbf{(A) } \dfrac{1}{3} \qquad$ $\textbf{(B) } \dfrac{\sqrt{2}}{2} \qquad$ $\textbf{(C) } \dfrac{3}{4} \qquad$ $\textbf{(D) } \dfrac{7}{9} \qquad$ $\textbf{(E) } \dfrac{\sqrt{3}}{2}$
Solution
$AC=\sqrt2$, assume $CD=x, AD=y$.
$x+y=2$, $y^2-x^2=2$, so $y-x=1$, $x=\frac12, y=\frac32$.
Right triangle $ACD$ with right angle at $C$ is constructed outward on the hypotenuse $\overline{AC}$ of the $30^\circ$-$60^\circ$-$90^\circ$ triangle $ABC$, which has its right angle at $B$, $\angle BAC=30^\circ$ and $BC=2$, so that the two triangles have equal perimeters. What is $CD$?
$\textbf{(A) } \dfrac{1}{3} \qquad$ $\textbf{(B) } \dfrac{\sqrt{2}}{2} \qquad$ $\textbf{(C) } \dfrac{3}{4} \qquad$ $\textbf{(D) } \dfrac{7}{9} \qquad$ $\textbf{(E) } \dfrac{\sqrt{3}}{2}$ 
