◀ ▲ ▶Branches / Geometry / Elements-euclid / Book--6-similar-figures / Proof: By Euclid

Proof: By Euclid

(related to Proposition: 6.28: Construction of Parallelogram Equal to Given Figure Less a Parallelogram)

• Let $AB$ have been cut in half at point $E$ [Prop. 1.10], and let (parallelogram) $EBFG$, (which is) similar, and similarly laid out, to (parallelogram) $D$, have been described on $EB$ [Prop. 6.18].
• And let parallelogram $AG$ have been completed.
• Therefore, if $AG$ is equal to $C$ then the thing prescribed has happened.
• For a parallelogram $AG$, equal to the given rectilinear figure $C$, has been applied to the given straight line $AB$, falling short by a parallelogrammic figure $GB$ which is similar to $D$.
• And if not, let $HE$ be greater than $C$.
• And $HE$ (is) equal to $GB$ [Prop. 6.1].
• Thus, $GB$ (is) also greater than $C$.
• So, let (parallelogram) $KLMN$ have been constructed (so as to be) both similar, and similarly laid out, to $D$, and equal to the excess by which $GB$ is greater than $C$ [Prop. 6.25].
• But, $GB$ [is] [similar]bookofproofs$1983 to $D$.
• Thus, $KM$ is also similar to $GB$ [Prop. 6.21].
• Therefore, let $KL$ correspond to $GE$, and $LM$ to $GF$.
• And since (parallelogram) $GB$ is equal to (figure) $C$ and (parallelogram) $KM$, $GB$ is thus greater than $KM$.
• Thus, $GE$ is also greater than $KL$, and $GF$ than $LM$.
• Let $GO$ be made equal to $KL$, and $GP$ to $LM$ [Prop. 1.3].
• And let the parallelogram $OGPQ$ have been completed.
• Thus, [$GQ$] is equal and similar to $KM$ [but, $KM$ is similar to $GB$].
• Thus, $GQ$ is also similar to $GB$ [Prop. 6.21].
• Thus, $GQ$ and $GB$ are about the same diagonal [Prop. 6.26].
• Let $GQB$ be their (common) diagonal, and let the (remainder of the) figure have been described.
• Therefore, since $BG$ is equal to $C$ and $KM$, of which $GQ$ is equal to $KM$, the remaining gnomon $UWV$ is thus equal to the remainder $C$.
• And since (the complement) $PR$ is equal to (the complement) $OS$ [Prop. 1.43], let (parallelogram) $QB$ have been added to both.
• Thus, the whole (parallelogram) $PB$ is equal to the whole (parallelogram) $OB$.
• But, $OB$ is equal to $TE$, since side $AE$ is equal to side $EB$ [Prop. 6.1].
• Thus, $TE$ is also equal to $PB$.
• Let (parallelogram) $OS$ have been added to both.
• Thus, the whole (parallelogram) $TS$ is equal to the gnomon $VWU$.
• But, gnomon $VWU$ was shown (to be) equal to $C$.
• Therefore, (parallelogram) $TS$ is also equal to (figure) $C$.
• Thus, the parallelogram $ST$, equal to the given rectilinear figure $C$, has been applied to the given straight line $AB$, falling short by the parallelogrammic figure $QB$, which is similar to $D$ [inasmuch as $QB$ is similar to $GQ$ " [Prop. 6.24] ":https://www.bookofproofs.org/branches/parallelograms-about-diameter-are-similar/\,].
• (Which is) the very thing it was required to do.
  ∎

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"