(iii) (b + c) (b + d) = (c + a) (c + d)
Solution:
(iii) (b + c) (b + d) = (c + a) (c + d)
According to the question, a, b, c are in GP.
Making use of the property of geometric mean, we can write:
b2 = ac
bc = ad
c2 = bd
Let us consider LHS: (b + c) (b + d)
$ =\left( b\text{ }+\text{ }c \right)\text{ }\left( b\text{ }+\text{ }d \right)\text{ } $
$ =\text{ }{{b}^{2}}~+\text{ }bd\text{ }+\text{ }cb\text{ }+\text{ }cd $
$ =\text{ }ac\text{ }+\text{ }{{c}^{2}}~+\text{ }ad\text{ }+\text{ }cd\text{ }\left[ by\text{ }using\text{ }property\text{ }of\text{ }geometric\text{ }mean \right] $
$ =\text{ }c\text{ }\left( a\text{ }+\text{ }c \right)\text{ }+\text{ }d\text{ }\left( a\text{ }+\text{ }c \right) $
$ =\text{ }\left( a\text{ }+\text{ }c \right)\text{ }\left( c\text{ }+\text{ }d \right) $
$ =\text{ }RHS $
∴ LHS = RHS
Hence proved.