Prove the following equation, given that $a,b,c,d$ are reals such that Given that the following is true for the real numbers $a,b,c,d$
$$\frac{a+b}{c+d}=\frac{a-b}{c-d}$$
Prove
$$(a^2 + b^2)(c^2 + d^2)=(ac + bd)^2$$
I cross multiplied the first equation and got $ad=bc$. Using this (substitution) I expanded and simplified the second equation to get
$$a^2 d^2=b^2 c^2$$
$$ad=bc$$
Which is the same as what I got before for the first equation. However, I feel like this is wrong because:


*

*I manipulate both equations. (I'm assuming that the second one is true)

*In the last step, by taking the square root of both sides I'm assuming that $ad,bc>0$


Is it wrong? Thanks.

$$\frac{a+b}{c+d}=\frac{a-b}{c-d}\implies \frac{a+b}{a-b}=\frac{c+d}{c-d}$$
Applying Componendo and dividendo, $$\frac{a+b+(a-b)}{a+b-(a-b)}=\frac{c+d+(c-d)}{c+d-(c-d)}\implies \frac ab=\frac cd\implies ad=bc$$
Now, $$(a^2+b^2)(c^2+d^2)=a^2c^2+b^2c^2+a^2d^2+b^2d^2=(ac+bd)^2+(ad-bc)^2$$ (There is another form :  Brahmagupta–Fibonacci identity)
$$\text{  As }ad=bc,(a^2+b^2)(c^2+d^2)=(ac+bd)^2 $$
