Question

daca a^2b+b^2c+c^2a=a^2c+b^2a+c^2b a triunghiului isoscel​

Answer 1

$\displaystyle\bf\\c)~a^2b+b^2c+c^2a=a^2c+b^2a+c^2b \Leftrightarrow ~a^2b+b^2c+c^2a - a^2c-b^2a-c^2b=0 \Leftrightarrow\\ daca~descompunem~in~factori~\implies vom~obtine~(a-b)(a-c)(b-c)=0,~implies\\a-b=0~sau~a-c=0~sau~b-c=0,~in~fiecare~caz~obtinem~:\\a=b~sau~a=c~sau~b=c,~de~unde~in~fiecare~caz~triunghiul~va~fi~isoscel.$

$\displaystyle\bf\\d)~da,~numerele~\sqrt{a},~\sqrt{b}~si~\sqrt{c}~pot~fi~lungimile~laturilor~unui~triunghi,~de\\exemplu~daca~a,b~si~c~sunt~patrate~perfecte~astfel~incat~sa~respecte~teroema\\lui~pitagora.$