Book 3
24
Therefore, if the straight line AB be applied to CD, the segment AEB will not fail to coincide with CFD also; therefore it will coincide with it and will be equal to it.
24
Therefore etc. Q. E. D. fall awry, παραλλάξει, the same word as used in the like case in I. 8. The word implies that the applied figure will partly fall short of, and partly overlap, the figure to which it is applied.
PROPOSITION 25.
25
Given a segment of a circle, to describe the complete circle of which it is a segment.
25
Let ABC be the given segment of a circle; thus it is required to describe the complete circle belonging to the segment ABC, that is, of which it is a segment.
25
For let AC be bisected at D, let DB be drawn from the point D at right angles to AC, and let AB. be joined; the angle ABD is then greater than, equal to, or less than the angle BAD.
25
First let it be greater; and on the straight line BA, and at the point A on it, let the angle BAE be constructed equal to the angle ABD; let DB be drawn through to E, and let EC be joined.
25
Then, since the angle ABE is equal to the angle BAE, the straight line EB is also equal to EA. [I. 6]
25
And, since AD is equal to DC, and DE is common, the two sides AD, DE are equal to the two sides CD, DE respectively; and the angle ADE is equal to the angle CDE, for each is right; therefore the base AE is equal to the base CE.
25
But AE was proved equal to BE; therefore BE is also equal to CE; therefore the three straight lines AE, EB, EC are equal to one another.
25
Therefore the circle drawn with centre E and distance one of the straight lines AE, EB, EC will also pass through the remaining points and will have been completed. [III. 9]
25
Therefore, given a segment of a circle, the complete circle has been described.
25
And it is manifest that the segment ABC is less than a semicircle, because the centre E happens to be outside it.
