If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjourning the equal angles, or that subtending one of the equal angles, they will have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.
[Desmos graphs can be found here]
So if two triangles have two pairs of equal angles and an equal side, we will prove that the remaining sides and angles are equal as well. This can be done by showing that the two triangles are equal.
Start with the two triangles ABC and DEF. Angle ABC is equal to angle DEF and angle BCA is equal to angle EFD.
Part 1: Let the sides adjoining the equal angles be equal, in other words, BC is equal to EF.
We will prove that AB is equal to DE, AC is equal to DF, and angle BAC is equal to angle EDF.
If AB is not equal to DE, let AB be greater than DE since one of them must be greater. Let G be a point on AB such that BG is equal to DE, and join CG.
We will compare triangle BCG to triangle DEF.
- BG is equal to DE (by the way we chose G)
- BC is equal to EF (those two sides were given to be equal)
- Angle GBC is equal to angle DEF (since angle GBC is equal to angle ABC)
Therefore, by Proposition 4 (side-angle-side), the base GC is equal to the base DF, the triangle BCG is equal to the triangle DEF, and angle GCB is equal to angle DFE.
But angle DFE is equal to angle ACB by hypothesis, which implies that angle GCB is equal to angle ACB. That is not possible. Thus side AB must be equal to side DE.
Returning to comparing triangles ABC and DEF …
We have the two sides AB and BC equal to the two sides DE and EF, and the angles contained between them (ABC and DEF) are equal.
Again using Proposition 4, the two bases AC and DF are equal and the remaining angles BAC and EDF are equal as well.
Part 2: Let two sides subtending equal angles be equal. We will choose that AB is equal to DE. (We could also choose that AC is equal to DF.)
We will prove that BC is equal to EF, AC is equal to DF, and angle BAC is equal to angle EDF.
If BC is not equal to EF, let BC be greater than EF since one of them must be greater. Let H be a point on BC such that BH is equal to EF, and join AH.
We will compare triangle ABH to triangle DEF.
- BH is equal to EF (by the way we chose H)
- AB is equal to DE (those two sides were given to be equal)
- Angle ABH is equal to angle DEF (since angle ABH is equal to angle ABC)
Once again, by Proposition 4 (side-angle-side), the base AH is equal to the base DF, the triangle ABH is equal to the triangle DEF, and angle AHB is equal to angle DFE.
But angle DFE is equal to angle ACB by hypothesis, which implies that angle AHB is equal to angle ACB. That is not possible. Thus side BC must be equal to side EF.
Returning to comparing triangles ABC and DEF …
We have the two sides AB and BC equal to the two sides DE and EF, and the angles contained between them (ABC and DEF) are equal.
By Proposition 4, the two bases AC and DF are equal, triangles ABC and DEF are equal, and the remaining angles BAC and EDF are equal as well.
Q.E.D.
Notice that the proofs in parts 1 and 2 are very similar.