Question

a²=b（b+c）,求证：A=2B

Answer 1

a²=b(b+c),
a²=b²+bc,
而,a=2R*sinA,b=2R*sinB,c=2R*sinC,
所以sin²A=sin²B+sinB●sinC,
sin²A-sin²B=sinB●sinC,
(sinA+sinB)(sinA-sinB)=sinB●sinC,
即
【2●
sin(A+B)/2
●
cos(A-B)/2】●【2
●
cos(A+B)/2
●
sin(A-B)/2】=sinB●sinC,
又因为2sin(A+B)/2
●cos(A+B)/2=sin(A+B)

2sin(A-B)/2●cos(A-B)/2=sin(A-B)
所以原式可化简为sin(A+B)●sin(A-B)=sinB●sinC
而,A+B+C=180,A+B=180-C,
sin(A+B)=sinC,
即有,sin(A-B)=sinB,
A-B=B,
A=2B,得证.