Euclid Elements Book 1 Proposition 26
If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equal angles, then the remaining sides equal the remaining sides and the remaining angle equals the remaining angle.
First part of proof (Given the side adjoining the equal angles are equal)
Prove: Triangle ABC = triangle DEF
Given BC=EF, angle B = angle E, angle C = angle F
To prove by contradiction:
Assume: AB > DE
1) Create point G such that BG = DE (Book I, Prop 3)
2) Create line GC (Postulate 1)
Now we have: BG = DE
BC = EF
angle B = angle E
So, if triangle GBC = triangle DEF (Book 1, Prop 4 )
3) Then angle GCB = angle DFE (Book 1, Prop 4)
Common Notion 1 states that things which equal the same thing are also equal one another. So,
4) angle GCB = angle ACB
But, this is not possible because of Common Notion 5, which states that the whole is greater than the part.
So, angle GCB cannot be equal to angle ACB. this makes the assumption of AB > DE not true.
Therefore, AB = DE
angle B = angle E
BC = EF
Which makes triangle ABC = triangle DEF (Book 1, Prop 4)
2nd part of proof, (Given the equal side is subtended from one of the equal angles)
Prove: triangle ABC = triangle DEF
Given BC=EF, angle A = angle D, angle C = angle F
To prove by contradiction:
Assume: AC > DF
1) Create point G such that CG = DF (Book I, Prop 3)
2) Create line BG (Postulate 1)
Now we have: angle BGC = angle EDF (Book 1, Prop 4)
BC = EF
angle BAC = angle DEF
So, if triangle GBC = triangle DEF (Book 1, Prop 4 )
3) Then angle BGC = angle EDF (Book 1, Prop 4)
Common Notion 1 states that things which equal the same thing are also equal one another. So,
4) angle BGC = angle BAC
But, angle BGC is an exterior angle to triangle ABG and Book 1, Prop 16 states that exterior angles are bigger than interior and opposite angles.
This leads to
5) angle BGC > angle BAC (Book i, Prop 16)
This is a contradiction to 4) as angle BGC cannot be both equal to and greater than angle BAC.
This makes our assumption of AC > DF not true.
Therefore, AC = DF
angle C = angle F
angle A = angle D
Which makes triangle ABC = triangle DEF (Book 1, Prop 4)