第4卷命题 5 · 作三角形外接圆

Let ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle ABC. 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. First let them meet within at F, and let FB, FC, FA be joined. 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.

设三角形ABC，分别作AB、AC的垂直平分线DF、EF，它们交于F。

[I. 4] 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. 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. Let it be circumscribed, as ABC. 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.

若F在三角形内，连接FA、FB、FC，由AD=DB、DF公共且垂直，得AF=BF；同理CF=AF，故FA=FB=FC。

Then, similarly, we shall prove that the point F is the centre of the circle circumscribed about the triangle ABC. 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. 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] 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.

若F在BC上，连接AF，同理可证F为外心。

Therefore about the given triangle a circle has been circumscribed. Q. E. F.

若F在三角形外，连接FA、FB、FC，同理可证FA=FB=FC，故以F为圆心、FA为半径的圆过A、B、C，即为外接圆。