$a,b,c,d$ are complex numbers corresponding to points $A,B,C,D$ lying on a circle with origin as center,and chord $AB⟂CD$. Find $ab+cd$ Question
Let on the Argand plane $a,b,c$ and $d$ represent the complex numbers corresponding to the points $A,B,C$ and $D$ respectively, all of which lie on a circle having center at the origin. The chord $AB$ is perpendicular to the chord $CD$.
Then find the value of $ab+cd$.
What I tried
I took $$a=x_1+iy_1$$ $$b=x_2+iy_2$$ $$c=x_3+iy_3$$ $$d=x_4+iy_4$$
I then found the complex numbers representing $AB$ and $CD$ and applied the condition that they are perpendicular. However that leads to other relations between $ac+bd$ and $ad+c$ and not $ab+cd$.
I figures that since the complex numbers lie on a circle, taking them in the form $a=e^{iθ}$ might be useful, but it lead to the some equations as before.
Any hints on how to solve the question are appreciated.
Thanks a lot in advance!
Regards
 A: 
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, $$ab+cd=r^2(\operatorname{cis}(\alpha+\beta)+\operatorname{cis}(\gamma+\delta))$$
$$=r^2(\operatorname{cis}(\alpha+\beta)+\operatorname{cis}(\pi+\alpha+\beta))$$
$$=r^2(\operatorname{cis}(\alpha+\beta)-\operatorname{cis}(\alpha+\beta))$$
$$=0$$
A: 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
$$\gamma+\delta=\alpha+\beta+180^{\circ}.$$
A: for the point A we use (1,0) in the Argand diagram. let b representing the point B have argument $\pi - 2\theta$ (with $0 \lt \theta \lt \frac{\pi}2$. the quadrilateral reads (anticlockwise) ACBD. (choosing b in the upper half plane is no loss of generality, by a symmetry argument).
Then the line AB makes an angle $\pi - \theta$ with OX measured in the usual anticlockwise direction, and the line DC, at right angles to AB, makes the angle $\frac{\pi}2 - \theta$ with OX.
suppose the argument of C is $\alpha$. then, since OD is the reflection of OC in a line through the origin parallel to AB we have:
$$
\arg d = 2 (\pi - \theta) - \alpha
$$
we now have:
$$
\arg ab = \arg b = \pi - 2 \theta
$$
and
$$
\arg cd =  \alpha + 2 (\pi - \theta) - \alpha = \arg ab + \pi
$$
or, since a,b,c,d are all of unit modulus,
$$ab = -cd $$
this gives the required result $$ab + cd = 0$$ rotation of the entire diagram to make A coincide with any chosen point on the unit circle merely multiplies $ab+cd$ by a factor of unit modulus, and hence it remains zero.
