+13 In triangle ABC, ∠C=90∘. A semicircle is constructed along side ¯AC that is tangent to ¯BC and ¯AB. If the radius of the semicircle is equal to BC3, then find ACBC.
+130466 
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
