if a,b,c,d are in continued proportion prove that:

(b-c)² (c-a)² +(d-b)² = (a-d)²

Answer :
We know when a , b ,c , d are in continued proportion , we get
$\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$
Let
$\frac{a}{b} = \frac{b}{c} = \frac{c}{d}$ = k  , Then  a  = bk   , b  = ck   And  c  = dk 
$\therefore$ b  =ck = dk.k   = dk^​2   , a  = bk  = dk².k   = dk³

$\therefore$ a = dk³  , b  = dk²  ,  c  = dk
We have to prove
( b -c )²  + ( c -a )²  + ( b - d )²   = ( a - d )²

Taking L.H.S. and substitute all values As :

$\Rightarrow$ ( dk² - dk )² + (  dk - dk³ )²  + ( dk² - d )²
$\Rightarrow$[ d²k⁴  + d²k²  - 2d²k³  ]  + [ d²k²  + d²k⁶  - 2d²k⁴  ] + [ d²k⁴ + d² - 2d²k² ]

$\Rightarrow$ d²k⁴ + d²k² - 2d²k³   +  d²k² + d²k⁶ - 2d²k⁴   + d²k⁴ + d² - 2d²k²

$\Rightarrow$d² + d²k⁶ - 2d²k³                ---------------- ( 1 )
Now taking R.H.S. And substitute values , we get
$\Rightarrow$( dk³ - d )²

$\Rightarrow$[ d²k⁶ + d²  - 2d²k³ ]

$\Rightarrow$d² + d²k⁶ - 2d²k³                ---------------- ( 2 )

Now we can see from equation 1 and 2 that

L.H.S.   = R.H.S.                                                   ( Hence proved  )