+13 In triangle $ABC,$ $\angle C = 90^\circ.$ A semicircle is constructed along side $\overline{AC}$ that is tangent to $\overline{BC}$ and $\overline{AB}.$ If the radius of the semicircle is equal to $\frac{BC}{3},$ then find $\frac{AC}{BC}.$
+129895 
Angle AED = 90 = Angle BED
DC = DE
DB = DB
Triangle DBC congrunent to triangle DBE by HL
tan (DBC) = DC / BC = (BC/3) / BC = 1/3 = tan (DBE)
tan (EBC) = AC / BC =
tan (2 * DBC) =
[ tan (DBE) + tan (DBC) ] / [ 1 - tan(DBE)*tan (DBC) ] =
(1/3 + 1/3 ) / ( 1 - (1/3)(1/3) ) =
(2/3) / ( 8/9) =
18/24 =
3/4 = AC / BC
