Intersecting Chord Theorem/Proof 2

Theorem
Let $CD$ and $EF$ both be chords of the same circle.

Let $CD$ and $EF$ intersect at $A$.

Then $CA \cdot AD = EA \cdot AF$.

Proof
Join $C$ with $F$ and $E$ with $D$, as shown in this diagram:



Then we have:

By Triangles with Two Equal Angles are Similar we have $\triangle FCA \sim \triangle DEA$.

Thus: