# Proof: By Euclid

• Let $BE$, $EC$, $GL$, and $LH$ have been joined.
• And since polygon $ABCDE$ is similar to polygon $FGHKL$, angle $BAE$ is equal to angle $GFL$, and as $BA$ is to $AE$, so $GF$ (is) to $FL$ [Def. 6.1] .
• Therefore, since $ABE$ and $FGL$ are two triangles having one angle equal to one angle and the sides about the equal angles proportional, triangle $ABE$ is thus equiangular to triangle $FGL$ [Prop. 6.6].
• Hence, (they are) also similar [Prop. 6.4], [Def. 6.1] .
• Thus, angle $ABE$ is equal to (angle) $FGL$.
• And the whole (angle) $ABC$ is equal to the whole (angle) $FGH$, on account of the similarity of the polygons.
• Thus, the remaining angle $EBC$ is equal to $LGH$.
• And since, on account of the similarity of triangles $ABE$ and $FGL$, as $EB$ is to $BA$, so $LG$ (is) to $GF$, but also, on account of the similarity of the polygons, as $AB$ is to $BC$, so $FG$ (is) to $GH$, thus, via equality, as $EB$ is to $BC$, so $LG$ (is) to $GH$ [Prop. 5.22], and the sides about the equal angles, $EBC$ and $LGH$, are proportional.
• Thus, triangle $EBC$ is equiangular to triangle $LGH$ [Prop. 6.6].
• Hence, triangle $EBC$ is also similar to triangle $LGH$ [Prop. 6.4], [Def. 6.1] .
• So, for the same (reasons), triangle $ECD$ is also similar to triangle $LHK$.
• Thus, the similar polygons $ABCDE$ and $FGHKL$ have been divided into equal numbers of similar triangles. I also say that (the triangles) correspond (in proportion) to the wholes.
• That is to say, the triangles are proportional: $ABE$, $EBC$, and $ECD$ are the leading (magnitudes), and their (associated) following (magnitudes are) $FGL$, $LGH$, and $LHK$ (respectively).
• (I) also (say) that polygon $ABCDE$ has a squared ratio to polygon $FGHKL$ with respect to (that) a corresponding side (has) to a corresponding side - that is to say, (side) $AB$ to $FG$.
• For let $AC$ and $FH$ have been joined.
• And since angle $ABC$ is equal to $FGH$, and as $AB$ is to $BC$, so $FG$ (is) to $GH$, on account of the similarity of the polygons, triangle $ABC$ is equiangular to triangle $FGH$ [Prop. 6.6].
• Thus, angle $BAC$ is equal to $GFH$, and (angle) $BCA$ to $GHF$.
• And since angle $BAM$ is equal to $GFN$, and (angle) $ABM$ is also equal to $FGN$ (see earlier), the remaining (angle) $AMB$ is thus also equal to the remaining (angle) $FNG$ [Prop. 1.32].
• Thus, triangle $ABM$ is equiangular to triangle $FGN$.
• So, similarly, we can show that triangle $BMC$ is also equiangular to triangle $GNH$.
• Thus, proportionally, as $AM$ is to $MB$, so $FN$ (is) to $NG$, and as $BM$ (is) to $MC$, so $GN$ (is) to $NH$ [Prop. 6.4].
• Hence, also, via equality, as $AM$ (is) to $MC$, so $FN$ (is) to $NH$ [Prop. 5.22].
• But, as $AM$ (is) to $MC$, so [triangle] $ABM$ is to $MBC$, and $AME$ to $EMC$.
• For they are to one another as their bases [Prop. 6.1].
• And as one of the leading (magnitudes) is to one of the following (magnitudes), so (the sum of) all the leading (magnitudes) is to (the sum of) all the following (magnitudes) [Prop. 5.12].
• Thus, as triangle $AMB$ (is) to $BMC$, so (triangle) $ABE$ (is) to $CBE$.
• But, as (triangle) $AMB$ (is) to $BMC$, so $AM$ (is) to $MC$.
• Thus, also, as $AM$ (is) to $MC$, so triangle $ABE$ (is) to triangle $EBC$.
• And so, for the same (reasons), as $FN$ (is) to $NH$, so triangle $FGL$ (is) to triangle $GLH$.
• And as $AM$ is to $MC$, so $FN$ (is) to $NH$.
• Thus, also, as triangle $ABE$ (is) to triangle $BEC$, so triangle $FGL$ (is) to triangle $GLH$, and, alternately, as triangle $ABE$ (is) to triangle $FGL$, so triangle $BEC$ (is) to triangle $GLH$ [Prop. 5.16].
• So, similarly, we can also show, by joining $BD$ and $GK$, that as triangle $BEC$ (is) to triangle $LGH$, so triangle $ECD$ (is) to triangle $LHK$.
• And since as triangle $ABE$ is to triangle $FGL$, so (triangle) $EBC$ (is) to $LGH$, and, further, (triangle) $ECD$ to $LHK$, and also as one of the leading (magnitudes is) to one of the following, so (the sum of) all the leading (magnitudes is) to (the sum of) all the following [Prop. 5.12], thus as triangle $ABE$ is to triangle $FGL$, so polygon $ABCDE$ (is) to polygon $FGHKL$.
• But, triangle $ABE$ has a squared ratio to triangle $FGL$ with respect to (that) the corresponding side $AB$ (has) to the corresponding side $FG$.
• For, similar triangles are in the squared ratio of corresponding sides [Prop. 6.14].
• Thus, polygon $ABCDE$ also has a squared ratio to polygon $FGHKL$ with respect to (that) the corresponding side $AB$ (has) to the corresponding side $FG$.
• Thus, similar polygons can be divided into equal numbers of similar triangles corresponding (in proportion) to the wholes, and one polygon has to the (other) polygon a squared ratio with respect to (that) a corresponding side (has) to a corresponding side.
• (Which is) the very thing it was required to show.

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