Question

cho a,b,clà độ dài ba cạnh tam giác

a^2 +b^2+c^2=ab+ac+bc.

c/m tam giac deu

Answer 1

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)

Answer 2

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