a2+b2+c2=ab+bc+ca
<=>2(a2+b2+c2)=2(ab+bc+ca)
<=>2a2+2b2+2c2=2ab+2bc+2ca
<=>2a2+2b2+2c2-2ab-2bc-2ca=0
<=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ca+a2)=0
<=>(a-b)2+(b-c)2+(c-a)2=0
Mà \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{matrix}\right.\)=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
=>\(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\) (đpcm)
vì \(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
Với mọi a,b,c ta luôn có \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\\left(a-c\right)^2\ge0\\\left(b-c\right)^2\ge0\end{matrix}\right.\)
mà \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
=> a-b=0 và b-c=0 và a-c=0
=> a=b=c
xét tam giác ABC có AB=AC=BC (a=b=c)
=> tam giác ABC đều
