Proposition I-24

Proposition 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 stright lines greater than the other, they will also have the base greater than the base.
Proof.

Since DF and DG are the same as AC the triangle DFG is isosceles.

Therefore the angles DFG and DGF are equal.

Let ABC and DEF be the given triangles with

AC equal to DF,
AB equal to DE, and angle at D greater than the angle at A.

We will show that the side FE is greater than the side CB.

Construct along DE an angle equal to the angle CAB.

Then construct on this angle a segment DG of the length of AC.

Join GE and GF

It follows that the angle EFG is less than the angle DGF, and consequently less than the angle FGE.

Thus the side FE is greater than the side FG, which is the same as BC.
(Prop I-18)

QED

(CN-x)

(Prop I-xx)

(Post-xx)