Root of Area contained by Rational Straight Line and First Binomial/Lemma

Proof

 * Euclid-X-54-Lemma.png

We have that $DB = BF$ and $BE = BG$.

Thus $DE = FG$.

But from :
 * $DE = AH = KC$

and:
 * $FG = AK = HC$.

Therefore:
 * $AH = AK = HC = KC$

Therefore the parallelogram $AKCH$ is equilateral.

The parallelogram $AKCH$ is also rectangular.

Therefore by definition $AKCH$ is a square.

We have that:
 * $FB : BG = DB : BE$

and:
 * $FB : BG = AB : DG$

and from :
 * $DB : BE = DG : BC$

it follows from :
 * $AB : DG = DG : BC$

Therefore $DG$ is a mean proportional between $AB$ and $BC$.

We also have that:
 * $AD : DK = KG : GC$

and so from :
 * $AK : KD = KC : CG$

while:
 * $AK : KD = AC : CD$

and from :
 * $KC : CG = DC : CB$

it follows from :
 * $AC : DC = DC : BC$

Therefore $DC$ is a mean proportional between $AC$ and $BC$.