Consider #DeltaABH#, by Pythagorean Theorem, #AB^2=AH^2+BH^2# #=> AB=sqrt(3^2+6^2)=sqrt45=3sqrt5#
As #DeltaABH and DeltaDBA# are similar, #=> (AD)/(AB)=(HA)/(HB)# #=> AD=(HA)/(HB)*AB=3/6*3sqrt5=(3sqrt5)/2#
Given #AD=DC, => AC=2AD=3sqrt5#
As #AC=AB, and angleA=90^@#, #=> angleABC=angleACB=45^@# #=> DeltaABC# is an isosceles right triangle. #=> BC=ABsqrt2=3sqrt5*sqrt2=3sqrt10#