Angles made by Chord with Tangent

Theorem
Let $EF$ be a tangent to a circle $ABCD$, touching it at $B$.

Let $BD$ be a chord of $ABCD$.

Then:
 * the angle in segment $BCD$ equals $\angle DBE$

and:
 * the angle in segment $BAD$ equals $\angle DBF$.

Proof

 * Euclid-III-32.png

Draw $BA$ perpendicular to $EF$ through $B$.

Let $C$ be selected on the circle on the arc $BD$.

Join $AD, DC, CB$.

From Right Angle to Tangent of Circle goes through Center, the center of the circle lies on $AB$.

By definition, then, $AB$ is a diameter of the circle.

From Relative Sizes of Angles in Segments, it follows that $\angle ADB$ is a right angle.

Therefore from Sum of Angles of Triangle Equals Two Right Angles $\angle BAD + \angle ABD$ equals a right angle.

But $\angle ABF$ is also a right angle.

So $\angle ABF = \angle BAD + \angle ABD$

Subtracting $\angle ABD$ from each, it follows that $\angle DBF = \angle BAD$.

Next we have that $ABCD$ is a cyclic quadrilateral.

From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles $\angle BAD + \angle BCD$ equals two right angles.

But from Two Angles on Straight Line make Two Right Angles, so does $\angle DBE + \angle DBF$.

But as $\angle BAD = \angle DBF$ it follows that $\angle BCD = \angle DBE$.