# Proof: By Euclid

Let each of the (angles) at $C$ and $F$ be assumed (to be) less than a right angle. * For if angle $ABC$ is not equal to (angle) $DEF$ then one of them is greater. * Let $ABC$ be greater. * And Let (angle) $ABG$, equal to (angle) $DEF$, have been constructed on the straight line $AB$ at the point $B$ on it [Prop. 1.23]. * And since angle $A$ is equal to (angle) $D$, and (angle) $ABG$ to $DEF$, the remaining (angle) $AGB$ is thus equal to the remaining (angle) $DFE$ [Prop. 1.32]. * Thus, triangle $ABG$ is equiangular to triangle $DEF$. * Thus, as $AB$ is to $BG$, so $DE$ (is) to $EF$ [Prop. 6.4]. * And as $DE$ (is) to $EF$, [so] it was assumed (is) $AB$ to $BC$. * Thus, $AB$ has the same ratio to each of $BC$ and $BG$ [Prop. 5.11]. * Thus, $BC$ (is) equal to $BG$ [Prop. 5.9]. * And, hence, the angle at $C$ is equal to angle $BGC$ [Prop. 1.5]. * And the angle at $C$ was assumed (to be) less than a right angle. * Thus, (angle) $BGC$ is also less than a right angle. * Hence, the adjacent angle to it, $AGB$, is greater than a right angle [Prop. 1.13]. * And ($AGB$) was shown to be equal to the (angle) at $F$. * Thus, the (angle) at $F$ is also greater than a right angle. * But it was assumed (to be) less than a right angle. * The very thing is absurd. * Thus, angle $ABC$ is not unequal to (angle) $DEF$. * Thus, (it is) equal. * And the (angle) at $A$ is also equal to the (angle) at $D$. * And thus the remaining (angle) at $C$ is equal to the remaining (angle) at $F$ [Prop. 1.32]. * Thus, triangle $ABC$ is equiangular to triangle $DEF$.

But, again, let each of the (angles) at $C$ and $F$ be assumed (to be) not less than a right angle. * I say, again, that triangle $ABC$ is equiangular to triangle $DEF$ in this case also. * For, with the same construction, we can similarly show that $BC$ is equal to $BG$. * Hence, also, the angle at $C$ is equal to (angle) $BGC$. * And the (angle) at $C$ (is) not less than a right angle. * Thus, $BGC$ (is) not less than a right angle either. * So, in triangle $BGC$ the (sum of) two angles is not less than two right angles. * The very thing is impossible [Prop. 1.17]. * Thus, again, angle $ABC$ is not unequal to $DEF$. * Thus, (it is) equal. * And the (angle) at $A$ is also equal to the (angle) at $D$. * Thus, the remaining (angle) at $C$ is equal to the remaining (angle) at $F$ [Prop. 1.32]. * Thus, triangle $ABC$ is equiangular to triangle $DEF$. * Thus, if two triangles have one angle equal to one angle, and the sides about other angles proportional, and the remaining angles both less than, or both not less than, right angles, then the triangles will be equiangular, and will have the angles about which the sides (are) proportional equal. * (Which is) the very thing it was required to show.

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