Consider each pair $(a, b)$ as the vector with the coordinates $(a, b)$ on the Cartesian plane. The inequality $ac+bd \ge 0$ is equivalent to the non-negativity of the scalar product, i.e. it states that the angle between the vectors is not greater than $\pi/2$. Also note that the lengths of the vectors don't matter, so without loss of generality assume that all these vectors are unit and their endpoints lie on the unit circle. Then for each fixed vector $(a, b)$ of the first set the inequality $ac+bd \ge 0$ means that the endpoint of the vector $(c, d)$ of the second set lies on the semicircle with the midpoint at the end of the vector $(a, b)$.

Thereby we can formulate the statement of the problem in the following equivalent form: Given the finite set $A$ of semicircles and the finite set $B$ of points on the unit circle such that every three semicircles from $A$ have a common point from $B$. Then all given semicircles from $A$ have a common point from $B$.

Let's prove the latter statement. We will identify each point on the unit circle with the oriented angle between the vector $\overrightarrow{(0,1)}$ and the radius-vector of the point, in particular this angle belongs to $[0, 2\pi)$. And fix the positive (anticlockwise) direction on the unit circle so that all semicircles from $A$ have unique starting point. Without loss of generality assume that one of the semicircles from $A$ starts at $0$ and ends at $\pi$. Consider the set of all semicircles from $A$ such that their starting point is greater than $\pi$, and let $[\ell_1, r_1]$ be the semicircle in this set with the smallest $r_1$, $\ell_1 > \pi > r_1$. Similarly consider the set of all semicircles from $A$ such that their starting point is less than $\pi$, and let $[\ell_2, r_2]$ be the semicircle in this set with the biggest $\ell_2$, $\ell_2 < \pi < r_2$. Since the semicircles $[0, \pi]$, $[\ell_1, r_1]$ and $[\ell_2, r_2]$ have a common point from $B$, there exists a given point $v \in B$ such that $r_1 \ge v \ge \ell_2$. Clearly, $v$ satisfies the statement conditions.

For the completeness of the proof let's make a few clarifications. Firstly, without loss of generality we assumed all semicircles in $A$ to be pairwise distinct, as well as all points $B$. Secondly, it could happen that it is impossible to find $[\ell_1, r_1]$ or $[\ell_2, r_2]$ since the corresponding set is empty. However, in these cases the assertion is obvious, since in them the desired point is $\pi$ or $0$, respectively. Finally, we ignore the semicircle $[\pi, 0]$; if such semicircle is given then the point $v$ cannot exist since otherwise the given semicircles $[\pi, 0]$, $[\ell_1, r_1]$ and $[\ell_2, r_2]$ don't have a common point from $B$; therefore the existence of $[\pi, 0]$ reduces to the previous case, and in this case it also contains a point common to all the others.