Book 4
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Let ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle ABC.
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Let the straight lines AB, AC be bisected at the points D, E [I. 10], and from the points D, E let DF, EF be drawn at right angles to AB, AC; they will then meet within the triangle ABC, or on the straight line BC, or outside BC.
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First let them meet within at F, and let FB, FC, FA be joined.
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Then, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base FB. [I. 4]
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Similarly we can prove that CF is also equal to AF; so that FB is also equal to FC; therefore the three straight lines FA, FB, FC are equal to one another.
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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.
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Let it be circumscribed, as ABC.
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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.
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Therefore about the given triangle a circle has been circumscribed. Q. E. F.
