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Proof: By Euclid

(related to Proposition: 1.45: Construction of Parallelograms III)

• Let $DB$ have been joined, and let the parallelogram $FH$, equal to the triangle $ABD$, have been constructed in the angle $HKF$, which is equal to $E$ [Prop. 1.42].
• And let the parallelogram $GM$, equal to the triangle $DBC$, have been applied to the straight line $GH$ in the angle $GHM$, which is equal to $E$ [Prop. 1.44].
• And since angle $E$ is equal to each of (angles) $HKF$ and $GHM$, (angle) $HKF$ is thus also equal to $GHM$.
• Let $KHG$ have been added to both.
• Thus, (the sum of) $FKH$ and $KHG$ is equal to (the sum of) $KHG$ and $GHM$.
• But, (the sum of) $FKH$ and $KHG$ is equal to two right angles [Prop. 1.29].
• Thus, (the sum of) $KHG$ and $GHM$ is also equal to two right angles.
• So two straight lines, $KH$ and $HM$, not lying on the same side, make adjacent angles with some straight line $GH$, at the point $H$ on it, (whose sum is) equal to two right angles.
• Thus, $KH$ is straight-on to $HM$ [Prop. 1.14].
• And since the straight line $HG$ falls across the parallels $KM$ and $FG$, the alternate angles $MHG$ and $HGF$ are equal to one another [Prop. 1.29].
• Let $HGL$ have been added to both.
• Thus, (the sum of) $MHG$ and $HGL$ is equal to (the sum of) $HGF$ and $HGL$.
• But, (the sum of) $MHG$ and $HGL$ is equal to two right angles [Prop. 1.29].
• Thus, (the sum of) $HGF$ and $HGL$ is also equal to two right angles.
• Thus, $FG$ is straight-on to $GL$ [Prop. 1.14].
• And since $FK$ is equal and parallel to $HG$ [Prop. 1.34], but also $HG$ to $ML$ [Prop. 1.34], $KF$ is thus also equal and parallel to $ML$ [Prop. 1.30].
• And the straight lines $KM$ and $FL$ join them.
• Thus, $KM$ and $FL$ are equal and parallel as well [Prop. 1.33].
• Thus, $KFLM$ is a parallelogram.
• And since triangle $ABD$ is equal to parallelogram $FH$, and $DBC$ to $GM$, the whole rectilinear figure $ABCD$ is thus equal to the whole parallelogram $KFLM$.
• Thus, the parallelogram $KFLM$, equal to the given rectilinear figure $ABCD$, has been constructed in the angle $FKM$, which is equal to the given (angle) $E$.
• (Which is) the very thing it was required to do.

fn1. The proof is only given for a four-sided figure. However, the extension to many-sided figures is trivial (translator's note).

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