Question

一直在△ABC中,三边长a,b,c满足等式a的平方-16b的平方-c的平方+6ab+10bc=0,求证a+c=2b

Answer 1

a^2-16b^2-c^2+6ab+10bc=0
a^2+6ab+9b^2-(25b^2-10bc+c^2)=0
(a+3b)^2-(5b-c)^2=0
a+3b=±(5b-c)
得：a+c=2b 或c-a=8b
又因为a、b、c是三角形三边，
所以c-a＜b
因此a+c=2b

Answer 2

a^2-16b^2-c^2+6ab+10bc=a^2+6ab+9b^2-(16b^2+9b^2-10bc+c^2)=(a+3b)^2-(5b-c)^2=0
(a+3b)^2=(5b-c)^2 a+3b=5b-c a+c=2b

Answer 3

a^2-16b^2-c^2+6ab+10bc=a^2+6ab+9b^2-(16b^2+9b^2-10bc+c^2)=(a+3b)^2-(5b-c)^2=0
(a+3b)^2=(5b-c)^2 a+3b=5b-c a+c=2b