# Proof: By Euclid

• For Let (angle) $FEG$, equal to angle $ABC$, and (angle) $EFG$, equal to $ACB$, have been constructed on the straight line $EF$ at the points $E$ and $F$ on it (respectively) [Prop. 1.23].
• Thus, the remaining (angle) at $A$ is equal to the remaining (angle) at $G$ [Prop. 1.32].
• Thus, triangle $ABC$ is equiangular to [triangle] $EGF$.
• Thus, for triangles $ABC$ and $EGF$, the sides about the equal angles are proportional, and (those) sides subtending equal angles correspond [Prop. 6.4].
• Thus, as $AB$ is to $BC$, [so] $GE$ (is) to $EF$.
• But, as $AB$ (is) to $BC$, so, it was assumed, (is) $DE$ to $EF$.
• Thus, as $DE$ (is) to $EF$, so $GE$ (is) to $EF$ [Prop. 5.11].
• Thus, $DE$ and $GE$ each have the same ratio to $EF$.
• Thus, $DE$ is equal to $GE$ [Prop. 5.9].
• So, for the same (reasons), $DF$ is also equal to $GF$.
• Therefore, since $DE$ is equal to $EG$, and $EF$ (is) common, the two (sides) $DE$, $EF$ are equal to the two (sides) $GE$, $EF$ (respectively).
• And base $DF$ [is] equal to base $FG$.
• Thus, angle $DEF$ is equal to angle $GEF$ [Prop. 1.8], and triangle $DEF$ (is) equal to triangle $GEF$, and the remaining angles (are) equal to the remaining angles which the equal sides subtend [Prop. 1.4].
• Thus, angle $DFE$ is also equal to $GFE$, and (angle) $EDF$ to $EGF$.
• And since (angle) $FED$ is equal to $GEF$, and (angle) $GEF$ to $ABC$, angle $ABC$ is thus also equal to $DEF$.
• So, for the same (reasons), (angle) $ACB$ is also equal to $DFE$, and, further, the (angle) at $A$ to the (angle) at $D$.
• Thus, triangle $ABC$ is equiangular to triangle $DEF$.
• Thus, if two triangles have proportional sides then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal.
• (Which is) the very thing it was required to show.

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