Intersecting Chord Theorem/Proof 2

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

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

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

Proof
Join $A$ with $B$ and $C$ with $D$, as shown in this diagram:


 * [[File:Euclid-III-35-2.png]]

Then we have:

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

Thus: