# Proof: By Euclid

• For let a third (straight line), $BG$, have been taken (which is) proportional to $BC$ and $EF$, so that as $BC$ (is) to $EF$, so $EF$ (is) to $BG$ [Prop. 6.11].
• And let $AG$ have been joined.
• Therefore, since as $AB$ is to $BC$, so $DE$ (is) to $EF$, thus, alternately, as $AB$ is to $DE$, so $BC$ (is) to $EF$ [Prop. 5.16].
• But, as $BC$ (is) to $EF$, so $EF$ is to $BG$.
• And, thus, as $AB$ (is) to $DE$, so $EF$ (is) to $BG$.
• Thus, for triangles $ABG$ and $DEF$, the sides about the equal angles are reciprocally proportional.
• And those triangles having one (angle) equal to one (angle) for which the sides about the equal angles are reciprocally proportional are equal [Prop. 6.15].
• Thus, triangle $ABG$ is equal to triangle $DEF$.
• And since as $BC$ (is) to $EF$, so $EF$ (is) to $BG$, and if three straight lines are proportional then the first has a squared ratio to the third with respect to the second [Def. 5.9] , $BC$ thus has a squared ratio to $BG$ with respect to (that) $CB$ (has) to $EF$.
• And as $CB$ (is) to $BG$, so triangle $ABC$ (is) to triangle $ABG$ [Prop. 6.1].
• Thus, triangle $ABC$ also has a squared ratio to (triangle) $ABG$ with respect to (that side) $BC$ (has) to $EF$.
• And triangle $ABG$ (is) equal to triangle $DEF$.
• Thus, triangle $ABC$ also has a squared ratio to triangle $DEF$ with respect to (that side) $BC$ (has) to $EF$.
• Thus, similar triangles are to one another in the squared ratio of (their) corresponding sides.
• (Which is) the very thing it was required to show.

Github: non-Github:
@Fitzpatrick