Question

if b^2=ac and 2b/1-b^2=a+c/1-ac then prove that a=b=c​

Answer 1

Answer:

Answer

Here the qua.dratic for third side b is given by

b

2

Answer 2

b2−2bccosA+(c2−a2)=0

∴b1+b2=2ccosA    ....(1)

and b1b2=c2−a2                ....(2)

Also it is given that 

b2=2b1                              ....(3)

Hence from (1) and (3),

3b1=2ccosA                           ....(4)

2b12=c2−a2                        ....(5)

Finally (4) and (5), we have

2.94c2cos2A=c2−a2

or 8c2(1−sin2A)=9c2−9a2

or 9a2=c2(1+8sin2A)

Hence 3a=c1+8sin2A.