# Intersecting Chord Theorem

## 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.

## Proof 1 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 $AE \cdot EC = DE \cdot EB$

$\blacksquare$

## Proof 2

Join $A$ with $B$ and $C$ with $D$, as shown in this diagram: Then we have:

 $\displaystyle \angle AEB$ $\cong$ $\displaystyle \angle DEC$ Two Straight Lines make Equal Opposite Angles $\displaystyle \angle BAE$ $\cong$ $\displaystyle \angle CDE$ Angles in Same Segment of Circle are Equal

By Triangles with Two Equal Angles are Similar we have $\triangle AEB \sim \triangle DEC$.

Thus:

 $\displaystyle \frac {AE} {EB}$ $=$ $\displaystyle \frac {DE} {EC}$ $\displaystyle \leadsto \ \$ $\displaystyle AE \cdot EC$ $=$ $\displaystyle DE \cdot EB$

$\blacksquare$

## Also known as

This result is also known as the chord theorem.