$a,b,c,d$ are real numbers such that.... 
Suppost that $a,b,c,d$ are real numbers such that $$a^2+b^2=1$$ $$c^2+d^2=1$$ $$ac+bd=0$$

I've to show that $$a^2+c^2=1$$  $$b^2+d^2=1$$ $$ab+cd=0$$
Basically,I've no any idea or tactics to tackle this problem. Any methods? Thanks in advance.
EDITED.
The given hint in the books is $S:=(a^2+c^2-1)^2+(b^2+d^2-1)^2=(ac+bd)^2$

Just for fun, here is a more advanced solution. Let
$g = \left( \begin{array}{cc} a & b \\
           c & d \end{array} \right)$
The equations in the hypothesis are equivalent to $gg^T=I_2$. The equations in the conclusion are equivalent to $g^Tg = I_2$. To get from the hypothesis to the conclusion, simply notice that $gg^T=I_2$ implies that $g^T = g^{-1}$.
Remark: in a more advanced language, the orthogonal group $O(n)$ is closed under transpose. This exercise is the $n=2$ special case.
