d(B',AD)=(T3p)=AB'=2√5 (1)
BB'=AA'=2√2 (2)
din (1) si (2) (Teo Pitagora inΔABB') ⇒AB=2√3
d(B',CD)=(T3p)=B'C =2√3 (3)
BB'=2√2 (2)
din(3) si (2), Teo Pitagora inΔBBC⇒BC=2
d(B',AC)=?
fie BP⊥AC, P∈AC
⇒(T3p)⇒d(B',AC)=B'P=√(BB'²+BP²)
BB'=2√2
BP = (inaltimea coresp ipotenuzei in ΔABC )=AB*BC/AC=2√3*2/4=√3
B'P=√(BB'²+BP²)=√((2√2)²+ (√3)²)=√(8+3)=√11
