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

Note that
$$0=(ac+bd)^2 = a^2c^2+b^2d^2+2abcd = a^2c^2+(1-a^2)(1-c^2) +2ac(-ac)$$
$$ = a^2c^2+1-a^2-c^2+a^2c^2-2a^2c^2 = 1-a^2-c^2$$
And so $a^2+c^2=1$. A similar argument shows that $b^2+d^2=1$. Further, now we can say
$$(ac+bd)^2-(ab+cd)^2 =a^2c^2+b^2d^2-a^2b^2-c^2d^2$$
$$ = a^2c^2+b^2d^2-a^2(1-d^2)-c^2(1-b^2)$$
$$=a^2c^2+b^2d^2+a^2d^2+b^2c^2-a^2-c^2$$
$$=(a^2+b^2)(c^2+d^2)-(a^2+c^2)=0$$
And so $$ab+cd=0$$.
