Tangent-Chord Theorem
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Theorem
Let $EF$ be a tangent to a circle $ABCD$, touching it at $B$.
Let $BD$ be a chord of $ABCD$.
Then:
and:
In the words of Euclid:
- If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line cutting the circle, the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle.
(The Elements: Book $\text{III}$: Proposition $32$)
Proof
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 $\angle DBF = \angle BAD$.
$\Box$
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 $\angle BCD = \angle DBE$.
$\blacksquare$
Also known as
The tangent-chord theorem is also known as the alternate segment theorem.
Historical Note
This proof is Proposition $32$ of Book $\text{III}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): circle $(7)$
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): tangent-chord theorem