设∠A,∠B,∠C成等差,其对边a,b,c成等比.
∵△ABC的三个内角成等差数列
∴2∠B=∠A+∠C
==>∠B=60
又∵b²=ac
由余弦定理得
b²=a²+c²-2accosB
ac=a²+c²-ac
(a-c)²=0
==>a=c
==>∠A=∠C=60
∴d=0
设∠A,∠B,∠C成等差,其对边a,b,c成等比.
∵△ABC的三个内角成等差数列
∴2∠B=∠A+∠C
==>∠B=60
又∵b²=ac
由余弦定理得
b²=a²+c²-2accosB
ac=a²+c²-ac
(a-c)²=0
==>a=c
==>∠A=∠C=60
∴d=0