A history of Greek mathematics - Wilbourhall.org

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,
the relation
AB.BD:AC.CD = BP 2 :CP 2 ;.
then, says Pappus, the ratio AP .
PD
:
BP
.
PC
be determined by
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.
Then, by hypothesis, AB .
BD
:
AC
.
CD
= BP 2 : CP
2 ;
2
,
therefore BF 2 : CG = BP 2 2 : OP
or
BF:CG = BP:CP,
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 . PG : BP PC
= ^4i J . Pi)
: £P
.
.
PC.