teorema Charles
in Δ ABM , AM = AB + BM = AB + BC /2 = ( 2AB + BC ) /2
in ΔADC , AN = AD + DN
DN / NC = 2 /1 proportii derivate ( adunam numaratorii la numitori)
DN / ( DN + NC) = 2 / ( 2 +1)
DN / DC = 2 /3 ⇒ DN = 2·DC / 3
AN = AD + 2·DC / 3 = ( 3AD + 2DC) / 3
DC = DA + AB + BC
AN = ( 3AD + 2DA + 2AB + 2BC) / 3
3AD + 2DA = 3AD - 2AD = AD
AN = ( AD + 2AB + 2BC) / 3