<s>[INST] $a,b,c,d$ 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 $S:=(a^2+c^2-1)^2+(b^2+d^2-1)^2=(ac+bd)^2$
 [/INST] Just a very basic brute force method,
$a^2 + b^2 = 1$                 - (1)
$c^2 + d^2 = 1$                  -(2)
$ac+bd=0$
On multiplying the two we get, $a^2c^2+b^2d^2+a^2d^2+b^2c^2=1$
Whereas $a^2c^2+b^2d^2+2abcd=0$ {squaring both sides $ac + bd = 0$}
Thus we get $-2abcd+a^2d^2+b^2c^2=1 \Rightarrow (ad-bc)^2=1 \Rightarrow ad-bc=\pm1$
Adding (1) and (2) we get, $a^2+b^2+c^2+d^2=2$ , Which can be re written as $a^2+b^2+c^2+d^2=2ad-2bc$.
$\Rightarrow (b+c)^2+(a-d)^2=0$. Thus we get, $b=-c$   and  $a =d$.
#Q.E.D
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