Intersecting Chord Theorem/Proof 1

Proof

 * Euclid-III-35.png

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 $AE \cdot EC = DE \cdot EB$