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

for the point A we use (1,0) in the Argand diagram. let b representing the point B have argument $pi - 2theta$ (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.

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 $a=roperatorname{cis}alpha$, $b=roperatorname{cis}beta$, $c=roperatorname{cis}gamma$ and $roperatorname{cis}delta$.

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

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 $DCperp AB$, $O$ and $CB$ are placed in the different sides respect to the line $AD$.

Also, let $ABcap CD={K}.$

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 $$gamma+delta=alpha+beta+180^{circ}.$$