Question

已知A^2+B^2+c^2-ab-bc-ca=0,请推导出A=B=C

Answer 1

a^2+b^2+c^2-ab-bc-ca=0
2a^2+2b^2+2c^2-2ab-2bc-2ca=0
a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ca+c^2=0
(a-b)^2+(b-c)^2+(a-c)^2=0
(a-b)^2=0,(b-c)^2=0,(a-c)^2=0
a=b.b=c,c=a
所扰烂哪以a=b=c
所以△ABC为缓码等边△
所历察以∠A=∠B=∠C

Answer 2

A^2+B^2+c^2-ab-bc-ca=0
2a^2+2b^2+2c^2-2ab-2bc-2ac=0
(a-c)^2+(b-c)^2+(a-b)^2=0
因为扰念|a-c|>=0
| b-c|>=0
|a-b|>=0
只有a-c=b-c=a-b=0的时缓仔困候戚配，等式才成立，即a=b=c

Answer 3

等式两边同乘2即 2(A^2+B^2+C^2-AB-BC-CA)=0 去毕码括号得运洞 2AA+2BB+2CC-2AB-2BC-2CA+0 结旁数枯合得（AA+BB-2AB)+(AA+CC-2AC)+(BB+CC-2BC)=0 即(A-B)^2+(A-C)^2+(B-C)^2=0 故A-B=0 A-C=0 B-C=0 因此A=B=C 得证