Angles made by Chord with Tangent

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

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


the angle in segment $BCD$ equals $\angle DBE$


the angle in segment $BAD$ equals $\angle DBF$.

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 $32$)



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 $\angle BCD = \angle DBE$.


Historical Note

This proof is Proposition $32$ of Book $\text{III}$ of Euclid's The Elements.