If a solid angle be contained by three plane angles, any two, taken together in any manner, are greater than the remaining one.
如果一个立体角由三个平面角构成，则任意两个平面角之和（无论怎样组合）都大于第三个平面角。
For let the solid angle at A be contained by the three plane angles BAC, CAD, DAB; I say that any two of the angles BAC, CAD, DAB, taken together in any manner, are greater than the remaining one. If now the angles BAC, CAD, DAB are equal to one another, it is manifest that any two are greater than the remaining one.
设立体角在点A由三个平面角BAC、CAD、DAB构成。若三个角相等，则显然任意两角之和大于第三角。
But, if not, let BAC be greater, and on the straight line AB, and at the point A on it, let the angle BAE be constructed, in the plane through BA, AC, equal to the angle DAB; let AE be made equal to AD, and let BEC, drawn across through the point E, cut the straight lines AB, AC at the points B, C; let DB, DC be joined. Now, since DA is equal to AE, and AB is common, two sides are equal to two sides; and the angle DAB is equal to the angle BAE; therefore the base DB is equal to the base BE.
若不等，设BAC最大。在直线AB上点A处，于平面BA、AC内作角BAE等于角DAB，取AE等于AD，过E作直线BEC交AB、AC于B、C，连接DB、DC。
[I. 4] And, since the two sides BD, DC are greater than BC, [I. 20] and of these DB was proved equal to BE, therefore the remainder DC is greater than the remainder EC. Now, since DA is equal to AE, and AC is common, and the base DC is greater than the base EC, therefore the angle DAC is greater than the angle EAC.
由于DA等于AE，AB公共，角DAB等于角BAE，故DB等于BE。又因BD、DC之和大于BC，且DB等于BE，故DC大于EC。
[I. 25] But the angle DAB was also proved equal to the angle BAE; therefore the angles DAB, DAC are greater than the angle BAC. Similarly we can prove that the remaining angles also, taken together two and two, are greater than the remaining one.
由于DA等于AE，AC公共，DC大于EC，故角DAC大于角EAC。又角DAB等于角BAE，所以角DAB与角DAC之和大于角BAC。同理可证其他组合。