$a,b,c,d$ 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 $S:=(a^2+c^2-1)^2+(b^2+d^2-1)^2=(ac+bd)^2$

What if $b=0?$
Else $abcd\ne0,$
we have $\dfrac ab=-\dfrac dc=k$(say)
$\implies a=bk,d=-ck$
$$1=a^2+b^2=b^2(1+k^2)\implies b^2=?,a^2=(bk)^2=?$$
$$1=c^2+d^2=c^2(1+k^2)\implies c^2=?,d^2=(-ck)^2=?$$
