Intersecting Chords Theorem/Proof 1

Theorem

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

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

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

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 $ABCD$.

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 $AE \cdot EC = DE \cdot EB$

$\blacksquare$