I am currently doing a complex number geometry course. I was tasked the following problem:

Problem Statement:
Let P be the point of intersection of the diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$. Further, let M be the midpoint of $AB$. Prove that $AC \perp BD$ if $\overleftrightarrow{PM} \perp CD$ and the center of circe in which $ABCD$ is inscribed doesn't lie on $PM$ Diagram 1

I am sorry in advance if my translation to the problem statement is a bit hard to read.

This problem looks very susceptible to complex bashing. But I am not quite sure how to carry out the calculations needed to establish the condition.

We employ complex numbers with the circumcircle of $\triangle ABC$ being the unit circle. Let
$A = a$, $M = \frac{a+b}{2}$
$B = b$, $P = \frac{ac(b+d) - bd(a+c)}{ac-bd}$
$C = c$
$D = d$

We further have that $\overleftrightarrow{PM} \perp CD$ $\Leftrightarrow$ $pm+cd = 0$ and we want to show that P is the midpoint of one of the diagonals. I am not quite sure how to do that exactly.

  • 1
    $\begingroup$ "Prove that AC⊥BC" : Is that "Prove that AC⊥BD" ? $\endgroup$
    – Prem
    Apr 3 at 11:49
  • $\begingroup$ Yes indeed i will edit that $\endgroup$
    – Helixglich
    Apr 3 at 11:51

1 Answer 1


Starting Hints ....

(A) Core Equality :
We have ,
(A1) : $ AC⊥BD \iff (c-a)/(d-b) = imaginary \iff (c-a)/(d-b) = - \overline {(c-a)/(d-b)} $

Likewise we have ,
(A2) : $ PM⊥CD \iff (m-p)/(d-c) = imaginary \iff (m-p)/(d-c) = - \overline {(m-p)/(d-c)} $

We have to start with (A2) , plug in the values , then simplify : In Case we get (A1) , we are Done.

(B) Values for ABCD : these are on the Unit Circle , hence $|z|=1$ & $\overline{z}=1/z$

(C) We can take values like $e^{i\theta_1}$ , $e^{i\theta_2}$ , $e^{i\theta_3}$ , $e^{i\theta_4}$ for the 4 Points ABCD.

(D) A Certain Division may involve some term in the Denominator which should not be Zero , hence the Criteria about the Center of the Circle.

I hope this helps ( "... I am not quite sure how to carry out the calculations needed ..." ) & you can Complete it now.

  • $\begingroup$ Uhhh I am not sure that steps A and B produce a solution. It is very computational heavy $\endgroup$
    – Helixglich
    Apr 3 at 13:12
  • $\begingroup$ Yes , it look heavy. If & when I get the time , I will attempt that , @Helixglich , Meantime , maybe somebody else will give a more Direct Answer ! $\endgroup$
    – Prem
    Apr 3 at 13:58

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