Intersecting Chords 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:

Then we have:

\(\ds \angle AEB\)

\(\cong\)

\(\ds \angle DEC\)

Two Straight Lines make Equal Opposite Angles

\(\ds \angle BAE\)

\(\cong\)

\(\ds \angle CDE\)

Angles in Same Segment of Circle are Equal

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

Thus:

\(\ds \frac {AE} {EB}\)

\(=\)

\(\ds \frac {DE} {EC}\)

\(\ds \leadsto \ \ \)

\(\ds AE \cdot EC\)

\(=\)

\(\ds DE \cdot EB\)

$\blacksquare$