$a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0$, compute $ab+cd$ 
Consider that $\{a,b,c,d\}\subset \Bbb R$ and it is known that
  $$a^2+b^2=c^2+d^2=1\ \ \text{and}\ \ ac+bd=0.$$
Compute the value of $ab+cd$. (Edit: originally written with a typo as $ab+bd$), 

This is from a list of training problems used within the preparation for the "Cone-Sul" math olympics. I tried to solve using newton identities, but without success. Any help will be appreciated. Sorry if this is a duplicate.
 A: $$(ab+cd)^2=(ab+cd)^2-(ac+bd)^2=a^2b^2+c^2d^2-a^2c^2-b^2d^2=$$
$$=(a^2-d^2)(b^2-c^2)=-(b^2-c^2)^2.$$
Thus, $$ab+cd=0.$$
A: If $ a^2 + b^2 = 1$, then $\exists \theta \in \mathbb{R}/ \cos \theta = a \text{ and } \sin \theta = b$.
Similarly, $ \exists \phi \in \mathbb{R}/ \cos \phi = c \text{ and } \sin \phi = d$.
Then, you can translate $ac+bd$ and $ab+bd$.
A: One approach is to  write the $\{a, b, c, d\}$ in terms of sines and cosines and utilise the angle sum formulae.
Thus write
\begin{align}
a &=\sin x \\
b &=\cos x\\
d&=\sin y \\
c &=\cos y 
\end{align}
and computing $ac+bd=\sin(x+y)$, also since $\cos(x+y)=bc-ad$ we obtain $$a^2+b^2=1,c^2+d^2=1\implies (ac+bd)^2+(ad-bc)^2=1$$
Alternatively, one could infer form complex numbers that $|z|=|w|=1\implies |zw|=1$. We can show more generally that the complex norm is multiplicative, of course. 
A: Let $a=\cos x,b=\sin x,c=\cos y,d=\sin y$. Then
$$ ac+bd=\cos x\cos y+\sin x\sin y=\cos(x-y) $$
and hence
$$ \cos(x-y)=0.$$
So
$$ ab+cd=\cos x\sin x+\cos y\sin y=\frac12(\sin(2x)+\sin(2y))=\sin(x+y)\cos(x-y)=0$$
A: Since 
$$ac+bd=0\implies a=-\frac{bd}{c}; c\neq0$$
Then
$$1=a^2+b^2=b^2[\frac{d^2}{c^2}+1]=b^2\cdot\frac{c^2+d^2}{c^2}=\frac{b^2}{c^2}$$
$$\implies c=\pm b\implies d=\mp a$$
The solution is given by
$$(a,b,c,d)=(a,b,\pm b,\mp a)$$
So
$$ab+cd=ab+(\pm b)(\mp a)=ab-ab=0$$
(For $c=0$, the case is trivial.)

Another solution
Let $a=mb$, $c=nd$.
$$ac+bd=0\implies mbnd+bd=0\implies bd=0 \text{ or }mn=-1$$
For $mn=-1$
$$a^2+b^2=b^2(m^2+1)=1\implies b^2=\frac{1}{m^2+1}$$
Similarly, $\displaystyle d^2=\frac{1}{n^2+1}$
Then
$$ab+cd=mb^2+nd^2=\frac{m}{m^2+1}+\frac{n}{n^2+1}=\frac{(m+n)(mn+1)}{(m^2+1)(n^2+1)}=0$$
A: Original: There is not information to compute $ab+bd$. Here are some examples that satisfy your conditions with different $ab+bd$:
$$ \begin{array}{cccc|c} a & b & c & d & ab+bd \\ \hline
1 & 0 & 1 & 0 & 0 \\
0.8 & 0.6 & -0.6 & 0.8 & 0.96 \\
0.8 & -0.6 & 0.6 & 0.8 & -0.96
\end{array} $$

After correction to question:
$ac+bd$ is the determinant of $$ \begin{pmatrix} a & -d \\ b & c \end{pmatrix} $$
where each column is a unit vector in $\mathbb R^2$ due to $a^2+b^2=c^2+d^2=1$. So the only way for this determinant to be $0$ is for the two columns to be either equal or each other's negative.
In the case $a=-d$, $b=c$ we have $ab+cd=ab-ab=0$.
In the case $a=d$, $b=-c$ we have $ab+cd=ab-ab=0$ again.
A: One has
$$ab+cd=ab(c^2+d^2)+cd(a^2+b^2)=(ac+bd)(ad+bc)=0.$$
Of course, the "right" proof of this is to note that the hypotheses
mean that
$$\pmatrix{a&b\\c&d}\pmatrix{a&c\\b&d}=\pmatrix{1&0\\0&1}$$
and since for square matrices, $AB=I$ implies $BA=I$ then
$$\pmatrix{a&c\\b&d}\pmatrix{a&b\\c&d}=\pmatrix{1&0\\0&1}$$
so that $ab+cd=0$ (and $a^2+c^2=b^2+d^2=1$).
