# Circle geometry. Prove quadrilateral diagonals are perpendicular.

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 $$PM$$

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.

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

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.

• Uhhh I am not sure that steps A and B produce a solution. It is very computational heavy Apr 3 at 13:12
• 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 !
– Prem
Apr 3 at 13:58