Book 3
22
But the angle CAB is equal to the angle BDC, for they are in the same segment BADC; [III. 21] and the angle ACB is equal to the angle ADB, for they are in the same segment ADCB; therefore the whole angle ADC is equal to the angles BAC, ACB.
22
Let the angle ABC be added to each; therefore the angles ABC, BAC, ACB are equal to the angles ABC, ADC. But the angles ABC, BAC, ACB are equal to two right angles; therefore the angles ABC, ADC are also equal to two right angles.
22
Similarly we can prove that the angles BAD, DCB are also equal to two right angles.
22
Therefore etc. Q. E. D.
PROPOSITION 23.
23
On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side.
23
For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB, ADB be constructed on the same side; let ACD be drawn through, and let CB, DB be joined.
23
Then, since the segment ACB is similar to the segment ADB, and similar segments of circles are those which admit equal angles, [III. Def. 11] the angle ACB is equal to the angle ADB, the exterior to the interior: which is impossible. [I. 16]
23
Therefore etc. Q. E. D. cannot be constructed, οὐ συσταθήσεται, the same phrase as in 1. 7.
PROPOSITION 24.
24
Similar segments of circles on equal straight lines are equal to one another.
24
For let AEB, CFD be similar segments of circles on equal straight lines AB, CD; I say that the segment AEB is equal to the segment CFD.
24
For, if the segment AEB be applied to CFD, and if the point A be placed on C and the straight line AB on CD, the point B will also coincide with the point D, because AB is equal to CD; and, AB coinciding with CD, the segment AEB will also coincide with CFD.
24
For, if the straight line AB coincide with CD but the segment AEB do not coincide with CFD, it will either fall with it, or outside it; or it will fall awry, as CGD, and a circle cuts a circle at more points than two : which is impossible. [III. 10]
