In an acute triangle ABC, let D be a point on [AC] and E be a point on [AB] such that \angle ADB\equal{}\angle AEC\equal{}90{}^\circ. If perimeter of triangle AED is 9, circumradius of AED is 59 and perimeter of triangle ABC is 15, then ∣BC∣ is<spanclass=′latex−bold′>(A)</span>5<spanclass=′latex−bold′>(B)</span>524<spanclass=′latex−bold′>(C)</span>6<spanclass=′latex−bold′>(D)</span>8<spanclass=′latex−bold′>(E)</span>548