Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$ $\textbf{(A) } 330 \qquad$ $ \textbf{(B) } 340 \qquad$ $ \textbf{(C) } 350 \qquad$ $ \textbf{(D) } 360 \qquad$ $ \textbf{(E) } 370$
Solution
Set $A=(0,0)$, $C=(20,0)$, $D=(20,30)$, and $E=(5,0)$. Since $\angle ABC=90^\circ$, point $B$ lies on the circle with diameter $AC$, so $x_B^2+y_B^2=20x_B$. Also, $B,D,E$ are collinear.
Write $B=E+t(E-D)=(5-15t,-30t)$. Substitution into the circle equation gives $15t^2+2t-1=0$. The root placing $B$ on the opposite side of $AC$ from $D$ is $t=1/5$, so $B=(2,-6)$.
The diagonals divide the quadrilateral into triangles $ABC$ and $ACD$ on opposite sides of $AC$. Their areas are $\frac12(20)(6)=60$ and $\frac12(20)(30)=300$, respectively.
Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=13,$ and $CD=12.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad$ $ \textbf{(B) } 340 \qquad$ $ \textbf{(C) } 350 \qquad$ $ \textbf{(D) } 360 \qquad$ $ \textbf{(E) } 370$ 
