已知在△ABC,三边长a、b、c满足等式a²-16b²-c²+6ab+10bc=0 求证:a+c=2b
问题描述:
已知在△ABC,三边长a、b、c满足等式a²-16b²-c²+6ab+10bc=0 求证:a+c=2b
答
a²-16b²-c²+6ab+10bc=0
a²+6ab+9b²-25b²-c²+10bc=0
(a²+6ab+9b²)-(25b²+c²-10bc)=0
(a+3b)²-(5b-c)²=0,
(a+3b)²=(5b-c)²,
所以a+3b=5b-c,或a+3b=-(5b-c),(舍去)
即a+c=2b