# Proof: By Euclid

• For Let (angle) $FDG$, equal to each of $BAC$ and $EDF$, and (angle) $DFG$, equal to $ACB$, have been constructed on the straight line $AF$ at the points $D$ and $F$ on it (respectively) [Prop. 1.23].
• Thus, the remaining angle at $B$ is equal to the remaining angle at $G$ [Prop. 1.32].
• Thus, triangle $ABC$ is equiangular to triangle $DGF$.
• Thus, proportionally, as $BA$ (is) to $AC$, so $GD$ (is) to $DF$ [Prop. 6.4].
• And it was also assumed that as $BA$ is) to $AC$, so $ED$ (is) to $DF$.
• And, thus, as $ED$ (is) to $DF$, so $GD$ (is) to $DF$ [Prop. 5.11].
• Thus, $ED$ (is) equal to $DG$ [Prop. 5.9].
• And $DF$ (is) common.
• So, the two (sides) $ED$, $DF$ are equal to the two (sides) $GD$, $DF$ (respectively).
• And angle $EDF$ [is] equal to angle $GDF$.
• Thus, base $EF$ is equal to base $GF$, and triangle $DEF$ is equal to triangle $GDF$, and the remaining angles will be equal to the remaining angles which the equal sides subtend [Prop. 1.4].
• Thus, (angle) $DFG$ is equal to $DFE$, and (angle) $DGF$ to $DEF$.
• But, (angle) $DFG$ is equal to $ACB$.
• Thus, (angle) $ACB$ is also equal to $DFE$.
• And (angle) $BAC$ was also assumed (to be) equal to $EDF$.
• Thus, the remaining (angle) at $B$ is equal to the remaining (angle) at $E$ [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 the equal angles proportional, 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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