Proposition: 1.26: "Angle-Side-Angle" and "Angle-Angle-Side" Theorems for the Congruence of Triangles

(Proposition 26 from Book 1 of Euclid's “Elements”)

If two triangles have two angles equal to two angles, respectively, and one side equal to one side - in fact, either that by the equal angles, or that subtending one of the equal angles - then (the triangles) will also have the remaining sides equal to the [corresponding] remaining sides, and the remaining angle (equal) to the remaining angle.

Let $ABC$ and $DEF$ be two triangles having the two angles $ABC$ and $BCA$ equal to the two (angles) $DEF$ and $EFD$, respectively. (That is) $ABC$ (equal) to $DEF$, and $BCA$ to $EFD$.
And let them also have one side equal to one side.
First of all, the (side) by the equal angles. (That is) $BC$ (equal) to $EF$.
I say that they will have the remaining sides equal to the corresponding remaining sides. (That is) $AB$ (equal) to $DE$, and $AC$ to $DF$.
And (they will have) the remaining angle (equal) to the remaining angle. (That is) $BAC$ (equal) to $EDF$.

fig26e

Modern Formulation

If two triangles ($\triangle{ABC}$, $\triangle{DEF}$) have two angles of one ($\alpha:=\angle{BAC}$, $\beta:=\angle{ACB}$) respectively equal to two angles of the other ($\gamma:=\angle{EDF}$, $\delta:=\angle{DFE}$), and a side of one equal to a similarly placed side of the other (placed with regard to the angles), then both triangles are congruent \[\triangle{ABC}\cong\triangle{DEF}.\]

This proposition breaks down into two cases according to whether the equal sides are adjacent or opposite to the equal angles: