An Identity in (Cyclic) Quadrilaterals

Problem

Let $ABCD$ be a cyclic quadrilateral; $E$ the intersection of the diagonals.

An Identity in (Cyclic) Quadrilaterals - problem

Then

$\displaystyle\frac{AE}{CE}=\frac{AB}{BC}\cdot\frac{AD}{CD}.$

Solution

Introduce the angles as shown in the diagram below$

An Identity in (Cyclic) Quadrilaterals

I shall apply the Law of Sines repeatedly in several triangles:

$\begin{align}\displaystyle \Delta ABE: & \frac{AE}{BE}=\frac{\sin\delta}{\sin\beta},\\ \Delta BCE: & \frac{BE}{CE}=\frac{\sin\gamma}{\sin\alpha},\\ \Delta ABC: & \frac{AB}{BC}=\frac{\sin\gamma}{\sin\beta},\\ \Delta ACD: & \frac{AD}{CD}=\frac{\sin\delta}{\sin\alpha}. \end{align}$

Multiplying the first and the second pair separately yields:

$\displaystyle \frac{AE}{CE}=\frac{AE}{BE}\frac{BE}{CE}=\frac{\sin\delta}{\sin\beta}\frac{\sin\gamma}{\sin\alpha}=\frac{AB}{BC}\frac{AD}{CD},$

which proves the statement.

Observation

The identity $\displaystyle\frac{AE}{CE}=\frac{AB}{BC}\cdot\frac{AD}{CD}$ is a necessary but not a sufficient condition for the quadrilateral $ABCD$ to be cyclic. The clue why this be so stems from its lack of symmetry. It compares poorly with the symmetry exhibited by Ptolemy's theorem that naturally comes to mind.

It is easy to exploit this lack of symmetry. Start with a square which is naturally cyclic:

An Identity in (Cyclic) Quadrilaterals - counterexample, part 1

Now, extend vertex $D$ along the diagonal $BD$ to a new position $D'.$ We still have $\displaystyle\frac{AE}{CE}=\frac{AB}{BC}\cdot\frac{AD'}{CD'}$ (because $\displaystyle\frac{AD}{CD}=1=\frac{AD'}{CD'})$ but the quadrilateral $ABCD'$ is no longer cyclic.

An Identity in (Cyclic) Quadrilaterals - counterexample, part 2

Acknowledgment

The problem has been posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page.

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