已知在△ABC中,三边a,b,c满足等式a²-16b²-c²+6ab+10bc=0,求证：a+c=2b

题目

已知在△ABC中,三边a,b,c满足等式a²-16b²-c²+6ab+10bc=0,求证：a+c=2b

答案

a²-16b²-c²+6ab+10bc=0
(a+3b)^2-(c-5b)^2=0
(a+3b+c-5b)(a+3b-c+5b)=0
(a+c-2b)(a+8b-c)=0
三角形种a+8b-c>0恒成立,所以a+c-2b=0
所以a+c=2b