If two triangles have the two sides equal to the 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.
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We start with two triangles ABC and DEF, for which side AB is equal to side DE, side AC is equal to side DF, and the angle at A is greater than the angle at D.
We will prove that side BC is greater than side EF.
Since angle BAC is greater than angle BAC, by Proposition 23 we can construct an angle EDG on the straight line DE that is equal to angle BAC.
(Proposition 23 covers the construction of an angle equal to a given angle.)
Choose the point G so that DG is equal to both AC and to DF. Join EG and FG.
Since AB is equal to DE and AC is equal to DG and angle BAC is equal to side EDG, then by Proposition 4 (side-angle-side) the base BC is equal to the base EG.
Since DF is equal to DG, angle DGF is equal to angle DFG by Proposition 5 (concerning equal angles in an isosceles triangle). Therefore angle DFG (being equal to angle DGF) is greater than angle EGF.
Since angle EFG is greater than angle DFG, then angle EFG is also greater than angle EGF.
Focus on triangle EFG.
Angle EFG is greater than angle EGF, so side EG is greater than side EF because Proposition 19 that the greater angle is subtended by the greater side.
But EG is equal to BC, so BC must also be greater than EF.
Q.E.D.
We are halfway through Book 1 – any comments are appreciated – George