# Intersecting Chord Theorem/Proof 1

## Contents

## Theorem

Let $AC$ and $BD$ both be chords of the same circle.

Let $AC$ and $BD$ intersect at $E$.

Then $AE \cdot EC = DE \cdot EB$.

In the words of Euclid:

*If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.*

(*The Elements*: Book $\text{III}$: Proposition $35$)

## Proof

Let $AC$ and $BD$ be intersecting chords of circle $ABCD$.

Let the point of intersection be $E$.

If $E$ is the center of $ABCD$ the solution is trivial, as $AE = EC = BE = ED$ and so $AE \cdot EC = BE \cdot ED$.

Otherwise, let $F$ be the center of $ABCD$.

Let $FG$ be drawn perpendicular to $AC$, and $FH$ be drawn perpendicular to $BD$.

From Conditions for Diameter to be Perpendicular Bisector, $G$ bisects $AC$ and $H$ bisects $BD$.

So $AG = GC$ and $BH = HD$.

From Difference of Two Squares we have that $AE \cdot EC + EG^2 = GC^2$.

Let us add $GF^2$ to these.

So $AE \cdot EC + EG^2 + GF^2 = GC^2 + GF^2$.

But from Pythagoras's Theorem we have that:

- $GC^2 + GF^2 = CF^2$
- $EG^2 + GF^2 = EF^2$

So:

- $AE \cdot EC + EF^2 = CF^2$

Using the same construction, we have that:

- $DE \cdot EB + EF^2 = BF^2$

But $BF = CF$ as both are the radius of the circle $ABCD$.

That gives us:

- $AE \cdot EC + EF^2 = DE \cdot EB + EF^2$

It follows that $AE \cdot EC = DE \cdot EB$

$\blacksquare$

## Historical Note

This theorem is Proposition $35$ of Book $\text{III}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions