$a,b,c,d$ 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

$g = \left( \begin{array}{cc} a &amp; b \\
           c &amp; d \end{array} \right)$

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.
It is boring. From $ac+bd=0$, $a=\frac{-bd}{c}$, put in $a^2+b^2=1$ the value of $a$, you have $\frac{b^2}{c^2}(d^2+c^2)=1$ now from $d^2+c^2=1$ you have $b^2=c^2$. similar arguments show that $a^2=d^2$. and together with $a=\frac{-bd}{c}$ you have all your results.

Put some effort to play with equations. Consider them all non-zero. If one is zero then obvious.

good luck