If there be three plane angles of which two, taken together in any manner, are greater than the remaining one, and they are contained by equal straight lines, it is possible to construct a triangle out of the straight lines joining the extremities of the equal straight lines.
若三个平面角中任意两个之和大于第三个，且它们由相等的线段所夹，则连接这些相等线段端点所得的线段可构成一个三角形。
Let there be three plane angles ABC, DEF, GHK, of which two, taken together in any manner, are greater than the remaining one, namely the angles ABC, DEF greater than the angle GHK, the angles DEF, GHK greater than the angle ABC, and, further, the angles GHK, ABC greater than the angle DEF; let the straight lines AB, BC, DE, EF, GH, HK be equal, and let AC, DF, GK be joined; I say that it is possible to construct a triangle out of straight lines equal to AC, DF, GK, that is, that any two of the straight lines AC, DF, GK are greater than the remaining one. Now, if the angles ABC, DEF, GHK are equal to one another, it is manifest that, AC, DF, GK being equal also, it is possible to construct a triangle out of straight lines equal to AC, DF, GK. But, if not, let them be unequal, and on the straight line HK, and at the point H on it, let the angle KHL be constructed equal to the angle ABC; let HL be made equal to one of the straight lines AB, BC, DE, EF, GH, HK, and let KL, GL be joined.
设三个平面角ABC、DEF、GHK，任意两角之和大于第三角，且边AB、BC、DE、EF、GH、HK相等，连接AC、DF、GK。
Now, since the two sides AB, BC are equal to the two sides KH, HL, and the angle at B is equal to the angle KHL, therefore the base AC is equal to the base KL. [I. 4] And, since the angles ABC, GHK are greater than the angle DEF, while the angle ABC is equal to the angle KHL, therefore the angle GHL is greater than the angle DEF.
若三个角相等，则AC、DF、GK也相等，显然可构成三角形。
And, since the two sides GH, HL are equal to the two sides DE, EF, and the angle GHL is greater than the angle DEF, therefore the base GL is greater than the base DF. [I. 24] But GK, KL are greater than GL. Therefore GK, KL are much greater than DF.
若角不等，在HK上作角KHL等于角ABC，取HL等于任一等边，连接KL、GL。由边角边得AC等于KL。
But KL is equal to AC; therefore AC, GK are greater than the remaining straight line DF. Similarly we can prove that AC, DF are greater than GK, and further DF, GK are greater than AC.
因角ABC与GHK之和大于角DEF，且角ABC等于角KHL，故角GHL大于角DEF；由边角边得GL大于DF。又GK加KL大于GL，故GK加KL大于DF，即AC加GK大于DF。同理可证其他两边之和大于第三边。