$a,b,c,d$ are complex numbers corresponding to points $A,B,C,D$ lying on a circle with origin as center,and chord $AB⟂CD$. Find $ab+cd$

Here, a is at an angle α, b is at an angle β, c is at an angle γ and d at an angle δ from the real number line. Given that AB perpendicular to CD, we can say that perpendicular bisectors of AB and CD are also perpendicular. So, We have :

Clearly we can say that |γ+ δ| + |α + β| = 180∘ or, γ + δ = α + β + 180∘

Now, $$ab+cd=r^2(\operatorname{cis}(\alpha+\beta)+\operatorname{cis}(\gamma+\delta))$$ $$=r^2(\operatorname{cis}(\alpha+\beta)+\operatorname{cis}(\pi+\alpha+\beta))$$ $$=r^2(\operatorname{cis}(\alpha+\beta)-\operatorname{cis}(\alpha+\beta))$$ $$=0$$

Let $a=r\operatorname{cis}\alpha$, $b=r\operatorname{cis}\beta$, $c=r\operatorname{cis}\gamma$ and $r\operatorname{cis}\delta$.

Thus, $\alpha+\beta=\gamma+\delta+180^{\circ}+360^{\circ}k,$ where $k\in\{-1,0\}$,

which says $$ab+cd=r^2(\operatorname{cis}(\alpha+\beta)+\operatorname{cis}(\gamma+\delta))=0.$$

For example, let $ABCD$ be our cyclic quadrilateral such that $DC\perp AB$, $O$ and $CB$ are placed in the different sides respect to the line $AD$.

Also, let $AB\cap CD=\{K\}.$

Thus, $$\measuredangle DCA=\measuredangle AKC+\measuredangle KAC$$ or $$\measuredangle DCA=90^{\circ}+\measuredangle BAC$$ or $$\frac{1}{2}(360^{\circ}-(\delta-\alpha))=90^{\circ}+\frac{1}{2}(\gamma-\beta)$$ or $$\gamma+\delta=\alpha+\beta+180^{\circ}.$$