# Proof: By Euclid

• For if $AB$ is unequal to $DE$ then one of them is greater.
• Let $AB$ be greater, and let $BG$ be made equal to $DE$ [Prop. 1.3], and let $GC$ have been joined.
• Therefore, since $BG$ is equal to $DE$, and $BC$ to $EF$, the two (straight lines) $GB$, $BC$1 are equal to the two (straight lines) $DE$, $EF$, respectively.
• And angle $GBC$ is equal to angle $DEF$.
• Thus, the base $GC$ is equal to the base $DF$, and triangle $GBC$ is equal to triangle $DEF$, and the remaining angles subtended by the equal sides will be equal to the (corresponding) remaining angles [Prop. 1.4].
• Thus, $GCB$ (is equal) to $DFE$.
• But, $DFE$ was assumed (to be) equal to $BCA$.
• Thus, $BCG$ is also equal to $BCA$, the lesser to the greater.
• The very thing (is) impossible.
• Thus, $AB$ is not unequal to $DE$.
• Thus, (it is) equal.
• And $BC$ is also equal to $EF$.
• So the two (straight lines) $AB$, $BC$ are equal to the two (straight lines) $DE$, $EF$, respectively.
• And angle $ABC$ is equal to angle $DEF$.
• Thus, the base $AC$ is equal to the base $DF$, and the remaining angle $BAC$ is equal to the remaining angle $EDF$ [Prop. 1.4].
• But, again, let the sides subtending the equal angles be equal: for instance, (let) $AB$ (be equal) to $DE$.
• Again, I say that the remaining sides will be equal to the remaining sides. (That is) $AC$ (equal) to $DF$, and $BC$ to $EF$.
• Furthermore, the remaining angle $BAC$ is equal to the remaining angle $EDF$.
• For if $BC$ is unequal to $EF$ then one of them is greater.
• If possible, let $BC$ be greater.
• And let $BH$ be made equal to $EF$ [Prop. 1.3], and let $AH$ have been joined.
• And since $BH$ is equal to $EF$, and $AB$ to $DE$, the two (straight lines) $AB$, $BH$ are equal to the two (straight lines) $DE$, $EF$, respectively.
• And the angles they encompass (are also equal).
• Thus, the base $AH$ is equal to the base $DF$, and the triangle $ABH$ is equal to the triangle $DEF$, and the remaining angles subtended by the equal sides will be equal to the (corresponding) remaining angles [Prop. 1.4].
• Thus, angle $BHA$ is equal to $EFD$.
• But, $EFD$ is equal to $BCA$.
• So, in triangle $AHC$, the external angle $BHA$ is equal to the internal and opposite angle $BCA$.
• The very thing (is) impossible [Prop. 1.16].
• Thus, $BC$ is not unequal to $EF$.
• Thus, (it is) equal.
• And $AB$ is also equal to $DE$.
• So the two (straight lines) $AB$, $BC$ are equal to the two (straight lines) $DE$, $EF$, respectively.
• And they encompass equal angles.
• Thus, the base $AC$ is equal to the base $DF$, and triangle $ABC$ (is) equal to triangle $DEF$, and the remaining angle $BAC$ (is) equal to the remaining angle $EDF$ [Prop. 1.4].
• Thus, 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.
• (Which is) the very thing it was required to show.

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### References

1. The Greek text has "$BG$, $BC$", which is obviously a mistake (translator's note).