△ABC的三个内角A,B,C的对边分别是a,b,c,求证:a^2=b(b+c)的充要条件是A=2B

Question

△ABC的三个内角A,B,C的对边分别是a,b,c,求证:a^2=b(b+c)的充要条件是A=2B

Answer 1

A = 2B => a^2 = b(b+c) :
做A的角平分线交BC于点D,
角BAD = 角CAD = 角B
因为 角B= 角CAD,角C = 角C,三角形ABC与三角形DAC相似.
所以,AD/BA = AC/BC AC/BC = CD/AC
AD * a = c*b 且 b^2 = a * CD
所以 b^2 + bc = a ( AD + DC ) = a^2
a^2 = b(b+c) => A = 2B
延长CA到点E,使AE = AB = c
因为 a^2 = b(b+c) 所以 a/(b+c) = b/a
又因为角C= 角C,所以△ABC与△BEC相似,
所以,∠ABC = ∠E,∠BAC = ∠EBC
又因为 AB = AE,所以,∠E = ∠ABE = ∠ABC
所以,∠BAC = ∠EBC = 2∠ABC,得证.