---
book: 3
number: 32
id: "III.32"
kind: "theorem"
uses: ["[[book-3/proposition-19]]", "[[book-3/proposition-31]]", "[[book-1/proposition-32]]", "[[book-3/proposition-22]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:3.prop.32"
license: "CC-BY-SA-4.0"
---

# III.32

*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*.

## Proof

For let a straight line *EF* touch the circle *ABCD* at the point *B*, and from the point *B* let there be drawn across, in the circle *ABCD*, a straight line *BD* cutting it; I say that the angles which *BD* makes with the tangent *EF* will be equal to the angles in the alternate segments of the circle, that is, that the angle *FBD* is equal to the angle constructed in the segment *BAD*, and the angle *EBD* is equal to the angle constructed in the segment *DCB*.

For let *BA* be drawn from *B* at right angles to *EF*, let a point *C* be taken at random on the circumference *BD*, and let *AD*, *DC*, *CB* be joined.

Then, since a straight line *EF* touches the circle *ABCD* at *B*, and *BA* has been drawn from the point of contact at right angles to the tangent, the centre of the circle *ABCD* is on *BA*. [[book-3/proposition-19|III. 19]]

Therefore *BA* is a diameter of the circle *ABCD*; therefore the angle *ADB*, being an angle in a semicircle, is right. [[book-3/proposition-31|III. 31]]

Therefore the remaining angles *BAD*, *ABD* are equal to one right angle. [[book-1/proposition-32|I. 32]]

But the angle *ABF* is also right; therefore the angle *ABF* is equal to the angles *BAD*, *ABD*.

Let the angle *ABD* be subtracted from each; therefore the angle *DBF* which remains is equal to the angle *BAD* in the alternate segment of the circle.

Next, since *ABCD* is a quadrilateral in a circle, its opposite angles are equal to two right angles. [[book-3/proposition-22|III. 22]]

But the angles *DBF*, *DBE* are also equal to two right angles; therefore the angles *DBF*, *DBE* are equal to the angles *BAD*, *BCD*, of which the angle *BAD* was proved equal to the angle *DBF*; therefore the angle *DBE* which remains is equal to the angle *DCB* in the alternate segment *DCB* of the circle.

Therefore etc. Q. E. D.