On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilinear angle.

Let *AB* be the given straight line, and the angle at *C* the given rectilinear angle.

It is required to describe on the given straight line *AB* a segment of a circle admitting an angle equal to the angle at *C.*

The angle at *C* is then acute, or right, or obtuse.

First let it be acute as in the first figure. Construct the angle *BAD* equal to the angle at *C* on the straight line *AB* and at the point *A.* Therefore the angle *BAD* is also acute.

Draw *AE* at right angles to *DA.* Bisect *AB* at *F.* Draw *FG* from the point *F* at right angles to *AB,* and join *GB.*

Then, since *AF* equals *FB,* and *FG* is common, the two sides *AF* and *FG* equal the two sides *BF* and *FG,* and the angle *AFG* equals the angle *BFG,* therefore the base *AG* equals the base *BG.*

Therefore the circle described with center *G* and radius *GA* passes through *B* also.

Draw it as *ABE,* and join *EB.*

Now, since *AD* is drawn from A, the end of the diameter *AE,* at right angles to *AE,* therefore *AD* touches the circle *ABE.*

Since then a straight line *AD* touches the circle *ABE,* and from the point of contact at A a straight line *AB* has been drawn across in the circle *ABE,* the angle *DAB* equals the angle *AEB* in the alternate segment of the circle.

But the angle *DAB* equals the angle at *C,* therefore the angle at *C* also equals the angle *AEB.*

Therefore on the given straight line *AB* the segment *AEB* of a circle has been described admitting the angle *AEB* equal to the given angle, the angle at *C.*

Next let the angle at *C* be right, and let it be again required to describe on *AB* a segment of a circle admitting an angle equal to the right angle at *C.*

Let the angle *BAD* be constructed equal to the right angle at *C,* as is the case in the second figure. Bisect *AB* at *F.* Describe the circle *AEB* with center *F* and radius either *FA* or *FB.*

Therefore the straight line *AD* touches the circle *ABE,* because the angle at A is right.

And the angle *BAD* equals the angle in the segment *AEB,* for the latter too is itself a right angle, being an angle in a semicircle.

But the angle *BAD* also equals the angle at *C,* therefore the angle *AEB* also equals the angle at *C.*

Therefore again the segment *AEB* of a circle has been described on *AB* admitting an angle equal to the angle at *C.*

Next, let the angle at *C* be obtuse.

Construct the angle *BAD* equal to *C* on the straight line *AB* and at the point A as is the case in the third figure. Draw *AE* at right angles to *AD.* Bisect *AB* again at *F.* Draw *FG* at right angles to *AB,* and join *GB.*

Then, since *AF* again equals *FB,* and *FG* is common, the two sides *AF* and *FG* equal the two sides *BF* and *FG,* and the angle *AFG* equals the angle *BFG,* therefore the base *AG* equals the base *BG.*

Therefore the circle described with center *G* and radius *GA* also passes through *B.* Let it so pass, as *AEB.*

Now, since *AD* is drawn at right angles to the diameter *AE* from its end, *AD* touches the circle *AEB.*

And *AB* has been drawn across from the point of contact at A, therefore the angle *BAD* equals the angle constructed in the alternate segment *AHB* of the circle.

But the angle *BAD* equals the angle at *C.*

Therefore the angle in the segment *AHB* also equals the angle at *C.*

Therefore on the given straight line *AB* the segment *AHB* of a circle has been described admitting an angle equal to the angle at *C.*

Q.E.F.

This proposition is not used in the rest of the *Elements.*