Proposition I.25 · Theorem
If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.
Drag the points circled in red. The figure changes, and every step of the proof still holds.
- Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, and let the base BC be greater than the base EF. The claim: the angle BAC is also greater than the angle EDF. (A, B, C, D can be dragged.)
- For if not, the angle BAC is either equal to the angle EDF or less.
- Now the angle BAC is not equal to the angle EDF, for then the base BC would be equal to the base EF; but it is not.Proposition I.4Proposition I.4: If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively.
- Neither is the angle BAC less than the angle EDF, for then the base BC would be less than the base EF; but it is not.Proposition I.24Proposition I.24: If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.
- Therefore the angle BAC is greater than the angle EDF.
Being what it was required to prove.