ΔABC
m(<A)=90 de grade
m(<B)=60 de grade ⇒m(<C)=180-150=30 de grade;
AD=6cm
[AD]=mediana ⇒AD=BC/2 ⇒BC=2·6=12cm;
Obtinem ca ΔADC-Δisoscel unde AD=DC=6cm
⇒m(<C)=m(<CAD)=30 de grade
⇒m(<ADC)=180-60=120 de grade;
m(<ADB)+m(<ADC)=180 de grade ⇒m(<ADB)=60 de grade;
Obtinem ca ΔBAD-Δechilateral unde AB=AD=BD.
Deci AB=6cm.