Let triangle ABC and triangle DEF be two triangles having the two sides line AB and line AC equal to the two sides line DE and line DF respectively, so that line AB equals line DE, and line AC equals line DF, and let the angle at A be greater than the angle at D.

I say that the base BC is greater than the base EF.

Since the angle BAC is greater than the angle EDF, construct the angle EDG equal to the angle BAC at the point D on the straight line DE. Make line DG equal to either of the two straight lines AC or DF. Join line EG and line FG.

Since line AB equals line DE, and line AC equals line DG, the two sides line BA and line AC equal the two sides line ED and line DG, respectively, and the angle BAC equals the angle EDG, therefore the base line BC equals the base line EG.

Again, since line DF equals line DG, therefore the angle DGF equals the angle DFG. Therefore the angle DFG is greater than the angle EGF.

Therefore the angle EFG is much greater than the angle EGF.

Since EFG is a triangle having the angle EFG greater than the angle EGF, and side opposite the greater angle is greater, therefore the side EG is also greater than line EF.

But line EG equals line BC, therefore line BC is also greater than line EF.

Therefore if two triangles have two sides equal to two sides respectively, but have one of the angles contained by the equal straight lines greater than the other, then they also have the base greater than the base.

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