(iv) (b-c)^2+(c-a)^2+(d-b)^2= (a-d)^2
I think you just want to show that this is true........???
a
b = ar
c = ar^2
d = ar^3
So
(b - c)^2 = (ar - ar^2)^2 = [ar (1 - r)]^2 = (ar)^2 (1 - r)^2 = (ar)^2 (1 - r)^2
(c - a)^2 = (ar^2 - a)^2 = [ a (r^2 - 1)]^2 = (a)^2 (r + 1)^2 (1 - r)^2
(d - b)^2 = ( ar^3 - ar)^2 = [ ar ( r^2 -1)]^2 = (ar)^2 (r + 1)^2 (1 - r)^2
(a - d)^2 = (a - ar^3)^2 = [ a (1 - r^3)]^2 = (a)^2 (1 - r)^2 ( 1 + r + r^2)^2
So
(b-c)^2+(c-a)^2+(d-b)^2= (a-d)^2
(ar)^2 (1 - r)^2 + (a)^2 (r + 1)^2 (1 - r)^2 + (ar)^2 (r + 1)^2 (1 - r)^2 = (a)^2 (1 - r)^2 ( 1 + r + r^2)^2
Divide through by a(1 - r)^2
r^2 + (r + 1)^2 + r^2 (1 + r)^2 = ( 1 + r + r^2) ^2
r^2 + [ r^2 + 2r + 1] + r^2 ( r^2 + 2r + 1) =
2r^2 + 2r + 1 + r^4 + 2r^3 + r^2 =
r^4 + 2r^3 + 3r^2 + 2r + 1 =
This factors as :
( 1 + r + r^2)^2
And this = the RHS
