Kitap 4
5
Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and the circle will have been circumscribed about the triangle ABC.
5
Let it be circumscribed, as ABC.
5
Next, let DF, EF meet on the straight line BC at F, as is the case in the second figure; and let AF be joined.
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Then, similarly, we shall prove that the point F is the centre of the circle circumscribed about the triangle ABC.
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Again, let DF, EF meet outside the triangle ABC at F, as is the case in the third figure, and let AF, BF, CF be joined.
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Then again, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base BF. [I. 4]
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Similarly we can prove that CF is also equal to AF; so that BF is also equal to FC; therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and will have been circumscribed about the triangle ABC.
5
Therefore about the given triangle a circle has been circumscribed. Q. E. F.
5
And it is manifest that, when the centre of the circle falls within the triangle, the angle BAC, being in a segment greater than the semicircle, is less than a right angle; when the centre falls on the straight line BC, the angle BAC, being in a semicircle, is right; and when the centre of the circle falls outside the triangle, the angle BAC, being in a segment less than the semicircle, is greater than a right angle. [III. 31]
