If $\frac ab= \frac bc= \frac cd$ then $(a^2+b^2)(c^2+d^2)=(ab+cd)^2$ [sic?] This is a basic algebra question. I found in class 9 math book and it is little tricky for me.

If $$\dfrac ab= \dfrac bc= \dfrac cd$$
  prove that  $$(a^2+b^2)(c^2+d^2)=(ab+cd)^2$$
Note (by @Blue). As observed in comments, the problem is incorrect as stated. It becomes valid if $(ab+cd)^2$ is replaced by $(ac+bd)^2$, but it is not clear if this is the source's intention. 


As noted in comments, the formula is wrong (or was, until I just corrected it).
Let $\alpha$ be the common ratio, so that $a=\alpha^3d,\ b=\alpha^2d,\ c=\alpha d$.  Then
$$(a^2+b^2)(c^2+d^2)=((\alpha^3d)^2+(\alpha^2d)^2)((\alpha d)^2+d^2)\\=(\alpha^6d^2+d^2)(\alpha^4d^2+d^2)=\alpha^4(\alpha^2d^2+d^2)(\alpha^2d^2+d^2)\\
=[\alpha^2(\alpha^2d^2+d^2)]^2=[\alpha^4d^2+\alpha^2d^2]^2\\
=[(\alpha^3d)(\alpha d)+(\alpha^2d)d]^2=(ac+bd)^2$$
