therefore the three angles CAB, ACB, ABC, are equal to the two angles ABC, ADC.
But the angles CAB, ACB, ABC are together equal to two right angles; [I. 32.
therefore also the angles ABC, ADC are together equal to two right angles.
In the same manner it may be shewn that the angles BAD, BCD are together equal to two right angles.
Wherefore, the opposite angles &c. q.e.d.
PROPOSITION 23. THEOREM.
On the same straight line, and on the same side of it, there cannot he two similar segments of circles, not coinciding with one another.
If it be possible, on the same straight line AB, and on the same side of it, let there be two similar segments of circles ACB, ADB, not coinciding with one another.
Then, because the circle ACB cuts the circle ADB at the two points A, B, they cannot cut one another at any other point; [III. 10.
therefore one of the segments must fall within the other; let ACB fall within ADB; draw the straight line BCD, and join AC, AD.
Then, because ACB, ADB are, by hypothesis, similar segments of circles, and that similar segments of circles contain equal angles, [III. Definition 11.
therefore the angle ACB is equal to the angle ADB;
that is, the exterior angle of the triangle ACD is equal to the interior and opposite angle;
which is impossible. [I. 16.
Wherefore, on the same straight line &c. q.e.d.
