If a,b,c,d are in continued proportion, prove that
(a−bc+a−cb)2−(d−bc+d−cb)2=(a−d)2(1c2+1b2)
Given a,b,c,d are in continued proportion
⟹ab=bc=cd=k(say)
⟹c=dk,b=ck=k2d,a=bk=k3d
LHS=(a−bc+a−cb)2−(d−bc+d−cb)2=(k3d−k2dkd+k3d−kdk2d)2−(d−k2dkd+d−kdk2d)2
=(k2−1k)2−(1k2−k)2=k4−1k4+1k2−k2+2k−2k
RHS=(a−d)2(1c2+1b2)=(k3−1)2(1k2+1k4)=k4−1k4+1k2−k2+2k−2k
LHS=RHS
Hence Proved