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 $EDF$.
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:
If the side in question is the side between the two angles, then the triangles are congruent.
If the side in question is not the side between the two angles, then the triangles are congruent.
respectively
Proofs: 1
Proofs: 1 2 3 4 5 6 7 8 9 10
Sections: 11