$a,b,c,d$ 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 $S:=(a^2+c^2-1)^2+(b^2+d^2-1)^2=(ac+bd)^2$
 A: 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.
A: Assume that $a$ and $b$ are the $\sin$ and $\cos$ of the angle $\alpha$, and $c$ and $d$ are the $\sin$ and $\cos$ of the angle $\beta$. Then $ac+bd=0$ gives you the relation betweeen $\alpha$ and $\beta$ to be used to prove what is required.
A: What if $b=0?$
Else $abcd\ne0,$
we have $\dfrac ab=-\dfrac dc=k$(say)
$\implies a=bk,d=-ck$
$$1=a^2+b^2=b^2(1+k^2)\implies b^2=?,a^2=(bk)^2=?$$
$$1=c^2+d^2=c^2(1+k^2)\implies c^2=?,d^2=(-ck)^2=?$$
A: 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 $$ab+cd=0$$.
A: 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
