fie BD bisectoarea ∡ABC, D∈AC
AC=√(20²-12²)=16
AD/DC=12/20=3/5 (teorema bisectoarei)⇒AD/AC=3/8⇒
⇒DC/AC=5/8
arie ΔBDC=(5/8)*Arie ΔABC=(5/8)*12*16/2=60cm²
arie ΔBDC=BD* d(C, BD) /2
60=BD* d(C,BD)/2
BD=√(AB²+AD²)=√(12²+6²)=6√5
60*2=6√5*d(C, BD)
d(C, BD)=120/6√5=20/√5=(20√5)/5=4√5
