Question

If a, b, c, d are in continued proportion, prove that:
$(b-c{)}^2+(c-a{)}^2+{(d-b)}^2=(a-d{)}^2$

Answer 1

$a,b,c,d$ are in continued proportion

$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}$

$\Rightarrow ac=b^2,bd=c^2,ad=bc$

consider

$(b-c{)}^2+(c-a{)}^2+(d-b{)}^2$

$\Rightarrow b^2+c^2-2bc+c^2+a^2-2ac+d^2+b^2-2bd$

$\Rightarrow a^2+2b^2+2c^2+d^2-2bc-2bd-2ac$

$\Rightarrow a^2+2\left(ac\right)+2\left(bd\right)+d^2-2bc-2bd-2ac$

$\Rightarrow a^2-2bc+d^2$

$\Rightarrow a^2-2bc+d^2=a^2-2\left(ad\right)+d^2$ since $ad=bc$

$=(a-d{)}^2$

Hence proved