Task B-1.4. (20 points) If $a^{2}+b^{2}=1, c^{2}+d^{2}=1$ and $a c+b d=0$, what is $a b+c d ?$
First solution. Since $a^{2}+b^{2}=1$, $a$ and $b$ cannot both be 0 at the same time. Let $a \neq 0$.

From the third equality, we have $c=-\frac{b d}{a}$. Then,

$$
1=c^{2}+d^{2}=\frac{b^{2} d^{2}}{a^{2}}+d^{2}=\frac{d^{2}\left(a^{2}+b^{2}\right)}{a^{2}}=\frac{d^{2}}{a^{2}}
$$

from which we have $a^{2}=d^{2}$. Now,

$$
a b+c d=a b-\frac{b d^{2}}{a}=\frac{b}{a}\left(a^{2}-d^{2}\right)=0
$$