262 THE DUPLICATION OF THE CUBE A
N ow MID = BM. MA + DA\ ai
while, by (1), FK 2 = BK. KG + OF 2 ; dl
therefore BM. if A + DA 2 = BK . KG + GF'\ tl
But DA — GF; therefore BM. MA — BK. KG. al
Therefore GK: MA = BM: BK L
= LG: GK;
F
while, at the same time, fh¥: J5ff = if A : AX. ^
Therefore LG:GK = GK: MA = MA : AL, t
or AB-.GK = GK: i/A = ifA : f?C.
1
{6) Apollonius, Heron, Philon of Byzantium.
I give these solutions* together because they really amount
to the same thing. 1
Let AB, AG, placed at right angles, be the two given straight
lines. Complete the rectangle ABDG, and let E be the point
at which the diagonals bisect one another.
Then a circle with centre E and radius EB will circumscribe
the rectangle ABDG.
Now (Apollonius) draw with centre E a circle cutting
AB, AC produced in F, G but such that F, D, G are in one
straight line.
Or (Heron) place a ruler so that its edge passes through D,
1 Heron’s solution is given in his Mechanics (i. 11) and Belopoeica, and is
reproduced by Pappus (hi, pp. 62-4) as well as by Eutocius (loc. cit.).