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.
 A: I would say it is not a good way, though sometimes works well, to assuming the equation you want to prove to be true in the first place.
What we usually do is to find the difference between the two expressions and see if the result is $0$.
Now, $(a^2+b^2)(c^2+d^2)-(ac+bd)^2=(a^2c^2+a^2d^2+b^2c^2+b^2d^2)-(a^2c^2+b^2d^2+2abcd)$
$=a^2d^2+b^2c^2-2abcd=abcd+abcd-2abcd=0$.
The last steps depends on the equality $ad=bc$ you get from the given condition.
A: An idea (and assuming that what has to be different from zero is and stuff):
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2\iff \color{red}{a^2c^2}+a^2d^2+b^2c^2+\color{blue}{b^2d^2}=\color{red}{a^2c^2}+2abcd+\color{blue}{b^2d^2}$$
$$\iff a^2d^2+b^2c^2=2abcd\iff (ad-bc)^2=0\iff ad=bc$$
Now the other condition:
$$\frac{a+b}{c+d}=\frac{a-b}{c-d}\iff \color{red}{ac}-ad+bc-\color{blue}{bd}=\color{red}{ac}+ad-bc-\color{blue}{bd}\iff$$
$$\iff 2ad=2bc\ldots$$
A: You need not (and must not) take the square root in the last step. You want to prove $a^2d^2 = b^2c^2$ and you already proved $ad=bc$. Simply regroup $a^2d^2$ to $(ad)^2$, apply the equation $ad=bc$ and simplify and there you are.
All in all, your reasoning is correct, but lacks some detail as to what you did precisely and where. I would prove the proposition is follows:


*

*Cross-multiply and simplify the assumption to obtain $ad=bc$, like you already did.

*Expand the left-hand side of the goal to $a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2$. Note that $a^2d^2=(ad)^2$, and we know from 1. that $ad=bc$. Therefore, this simplifies to $a^2c^2 + 2b^2c^2 + b^2d^2$

*Expand the right-hand side of the goal to $a^2c^2 + 2acbd + b^2d^2$. Note that $2acbd = 2(ad)(bc)$ and, again, because we know from 1. that $ad=bc$, this is $2(bc)^2$ and therefore the entire expression simplifies to $a^2c^2 + 2b^2c^2 + b^2d^2$


In conclusion, we have reduced the left-hand side of the goal and the right-hand side to the same expression, which is a rigorous proof.



In response to Shu Xiao Li's answer, I would also like to point out that it is not a good idea to assume the equation you are to show holds. This is not sound reasoning.
A: $$\frac{a+b}{c+d}=\frac{a-b}{c-d}\implies \frac{a+b}{a-b}=\frac{c+d}{c-d}$$
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)
$$\text{  As }ad=bc,(a^2+b^2)(c^2+d^2)=(ac+bd)^2 $$
