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

First let it be acute, and, as in the first figure, on the straight line AB , and at the point A , let the angle BAD be constructed equal to the angle at C ; therefore the angle BAD is also acute.

Let AE be drawn at right angles to DA , let AB be bisected at F , let FG be drawn from the point F at right angles to AB , and let GB be joined.

Then, since AF is equal to FB , and FG is common, the two sides AF , FG are equal to the two sides BF , FG ; and the angle AFG is equal to the angle BFG ; therefore the base AG is equal to the base BG . [[I. 4](elem.1.4)]

Therefore the circle described with centre G and distance GA will pass through B also.

Let it be drawn, and let it be ABE ; let EB be joined.

Now, since AD is drawn from A , the extremity of the diameter AE , at right angles to AE , therefore AD touches the circle ABE . [[III. 16, Por.](elem.3.16.p.1)]

Since then a straight line AD touches the circle ABE , and from the point of contact at A a straight line AB is drawn across in the circle ABE , the angle DAB is equal to the angle AEB in the alternate segment of the circle. [[III. 32](elem.3.32)]

But the angle DAB is equal to the angle at C ; therefore the angle at C is also equal to 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; let AB be bisected at F , and with centre F and distance either FA or FB let the circle AEB be described.

Therefore the straight line AD touches the circle ABE , because the angle at A is right. [[III. 16, Por.](elem.3.16.p.1)]

And the angle BAD is equal to the angle in the segment AEB , for the latter too is itself a right angle, being an angle in a semicircle. [[III. 31](elem.3.31)]

But the angle BAD is also equal to the angle at C .

Therefore the angle AEB is also equal to 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;

and on the straight line AB , and at the point A , let the angle BAD be constructed equal to it, as is the case in the third figure; let AE be drawn at right angles to AD , let AB be again bisected at F , let FG be drawn at right angles to AB , and let GB be joined.

Then, since AF is again equal to FB , and FG is common, the two sides AF , FG are equal to the two sides BF , FG ; and the angle AFG is equal to the angle BFG ; therefore the base AG is equal to the base BG . [[I. 4](elem.1.4)]

Therefore the circle described with centre G and distance GA will pass through B also; let it so pass, as AEB .

Now, since AD is drawn at right angles to the diameter AE from its extremity, AD touches the circle AEB . [[III. 16, Por.](elem.3.16.p.1)]

And AB has been drawn across from the point of contact at A ; therefore the angle BAD is equal to the angle constructed in the alternate segment AHB of the circle. [[III. 32](elem.3.32)]

But the angle BAD is equal to the angle at C .

Therefore the angle in the segment AHB is also equal to 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.