1985 AMC 12 第 22 题

The $60^\circ$ arc $CD$ makes $\angle CAD=30^\circ$

Use the resulting $30^\circ$-$60^\circ$-$90^\circ$ triangles to compare $AB$ and $AC$

Arc $CD$ gives $\angle CAD=30^\circ.$ Since $A,B,C$ are collinear and $A,O,D$ are collinear, $\angle BAO=30^\circ.$ With $\angle ABO=60^\circ,$ triangle $ABO$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle. Because $BO=5,$ we get $AB=10$ and $AO=5\sqrt3.$

Also $\angle ACD=90^\circ$ because $AD$ is a diameter. In the $30^\circ$-$60^\circ$-$90^\circ$ triangle $ACD,$ the hypotenuse is $AD=10\sqrt3,$ so $AC=15.$ Therefore $BC=AC-AB=5.$

Thus the correct answer is D.