△ABC三边a,b,c 满足a2+b2+c2 =ab+bc+ca,试判定△ABC的形状
△ABC三边a,b,c 满足a2+b2+c2 =ab+bc+ca,试判定△ABC的形状
△ABC三边a,b,c 满足a2+b2+c2 =ab+bc+ca,试判定△ABC的形状
△ABC三边a,b,c 满足a2+b2+c2 =ab+bc+ca,试判定△ABC的形状
a2+b2+c2 =ab+bc+ca
=> 2a2+2b2+2c2 =2ab+2bc+2ca
=> a2+b2-2ab+b2+c2-2bc+c2+a2-2ca=0
=>(a-b)2+(b-c)2+(c-a)2=0
=> a-b = 0,b-c=0,c-a=0
=>a=b=c
△ABC为等边三角形
两边同时*2,将右边所有项移到左边得
a2-2ab+b2+a2-2ac+c2+b2-2bc+c2=0
(a-b)2+(a-c)2+(b-c)2=0
because (a-b)2,(a-c)2,(b-c)2>=(大于或等于)0
so a-b=0 a-c=0 b-c=0
即a=b=c
所以是等边三角形
原等式化为:
2a2+2b2+2c2 =2ab+2bc+2ca
a2-2ab+b2+a2-2ac+c2+b2-2bc+c2=0
(a-b)^2+(a-c)^2+(b-c)^2=0
(a-b)^2≥0 (a-c)^2≥0 (b-c)^2≥0
所以(a-b)^2=(a-c)^2=(b-c)^2=0
所以a=b=c
三角形为等边三角形
我倒,慢了两楼
两边同时乘二 把左边的移过来
得:(a2-2ab+b2)+(a2-2ac+c2)+(b2-2bc+c2)=0
所以:(a-b)2+(a-c)2+(b-c)2=0
根据平方的性质
所以:a=b=c
所以为等边三角形
a^2+b^2+c^2 =ab+bc+ca
两边同时乘以2
2a^2+2b^2+2c^2 =2ab+2bc+2ca
a^2+b^2+a^2+c^2++b^2 +c^2=2ab+2bc+2ca
a^2+b^2-2ab+a^2+c^2-2bc+b^2 +c^2-2ca=0
(a-b ^2)+(b-c )^2+(a-c )^2=0
所以a=b=c
即为等边三角形
这题目我做过