Parallel Chords Cut Equal Chords in a Circle

Theorem

Let $\CC$ be a circle on center $O$.

Let $AB$ and $CD$ be chords in $\CC$ with $AB \parallel CD$.

Then the two chords cut from the circle by $AB$ and $CD$ are equal:

$AC = BD$

Proof

Draw $AD$.

By Parallelism implies Equal Alternate Angles:

$\angle BAD = \angle ADC$

The central angles corresponding to $\angle BAD$ and $\angle ADC$ are equal by Inscribed Angle Theorem:

$\angle BOD = \angle AOC$

As radii of $\CC$:

$OB = OD = OC = OA$

By Triangle Side-Angle-Side Congruence:

$\triangle OAC \cong \triangle OBD$

So:

$AC = BD$

$\blacksquare$