How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?

HINT:
$$ac+bd=0\iff\dfrac ad=\dfrac b{-c}=\pm\sqrt{\dfrac{a^2+b^2}{d^2+(-c)^2}}$$
But if $a^2+b^2=d^2+c^2,$ not necessarily $=1$
$$\dfrac ad=\dfrac b{-c}=\pm1$$
So, either $a=d,b=-c$ or $a=-d,b=c$
The result should follow immediately.
