A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

PG
THE COLLECTION. BOOK VII 411
Now suppose (A, D), (B, C) to be two point-pairs on a
straight line, and let P, another point on it, be determined by
the relation
AB.BD:AC.CD = BP 2 :CP 2 ;.
then, says Pappus, the ratio AP .
PD
:
BP
.
PC
is singular and
a minimum, and is equal to
AD 2 : ( VAC.BD- VTbTCD) 2 .
On iDas diameter draw a circle, and draw BF, CG perpendicular
to AD on opposite sides.
412 PAPPUS OF ALEXANDRIA
Therefore
AP.PD.BP .PC = AD 2 : { V(AC .
BD)- V{AB .
CD)}*.
The proofs of Props. 62 and 64 are different, the former
being long and involved. The results are :
Prop. 62.
If P is between C and D, and
AD.DB:AC.CB = DP 2 : PG\
then the ratio AP . PB : CP . PD is singular and a minimum
2
and is equal to { V(AC . BD) + V(AD . BC) : }
DC
Prop. 64.
If P is on AD produced, and
2 .
then the ratio AP . Pi)
AB.BD:AC.CD = BP 2 : CP
.
: 5P PC
and is equal to AD 2 :
{ V^C. BD) + V(AB .
2 ,
is singular and a maximum,
CD)}
2 .
,
(y) Lemmas on the Nevveis of Apollonius.
Then, by hypothesis, AB .
BD
:
AC
.
therefore BF 2 : CG = BP 2 2 : OP
or
BF:CG = BP:CP,
CD = BP 2 : CP
2
,
2 ;
whence the triangles FBP, GCP are similar and therefore
equiangular, so that FPG is a straight line.
Produce GO to meet the circle in H, join FH, and draw DK
perpendicular to FH produced. Draw the diameter FL and
join LH.
Now, by the lemma, FK = AC 2 . BD, and HK 2 = AB.CD;
therefore FH = FK - HK = V(AC . BD) - V(AB . CD).
Since, in the triangles FHL, PCG, the angles at H, C are
right and Z FLH= /.PGC, the triangles are similar, and
GP:PC = FL:FH = AD: FH
= AD: {V(AC.BD)- V(AB.CD)}.
But
GP:1 J C=FP:PB;
therefore GP 2 : PC = FP 2 . : BP . PC
= ^4i J . Pi)
: £P
.
PC.
After a few easy propositions (e.g. the equivalent of the
proposition that, if ax + x 2 = by + y
2
, then, according as a >
or < b, a + x > or < b + y), Pappus gives (Prop. 70) the
lemma leading to the solution of the vevais with regard to
the rhombus (see pp. 190-2, above), and after that the solution
by one Heraclitus of the same problem with respect to
a square (Props. 71, 72, pp. 780-4). The problem is, Given a
square A BCD, to draw through B a straight line, meeting CD
in H and AD produced in E, such that HE is equal to a given
length.
The solution depends on a lemma to the effect that, if any
straight line BHE through B meets CD in H and AD produced
in E, and if EF be drawn perpendicular to
BC produced in F, then
CF 2 = BC 2 +HE 2 .
BE meeting