THE COLLECTION. BOOK VII
411
Now suppose (A, D), (B, G) to be two point-pairs on a
straight line, and let P, another point on it, be determined by
the relation
AB.BD-.AC.GD = BP 2 : CP 2 ;
then, says Pappus, the ratio AP . PD : BP . PC is singular and
a minimum, and is equal to
AD 2 : (VAG.BD- VAB . CD) 2 .
On AD as diameter draw a circle, and draw BF, CG perpen
dicular to AD on opposite sides.
Then, by hypothesis, AB . BD : AC . CD = BP 2 : CP 2 ;
therefore BF 2 : CG 2 = BP 2 : CP 2 ,
or BF : CG = BP : CP,
whence the triangles FBP. GGP are similar and therefore
equiangular, so that FPG is a straight line.
Produce GG to meet the circle in H, join FH, and draw DK
perpendicular to FH produced. Draw the diameter PP and
join LH.
Now, by the lemma, FK 2 = AG. BD, and HK 2 = AB. CD ;
therefore FH = FK - HK = V{AG. BD) - V{AB . CD).
Since, in the triangles FHL, PGG, the angles at H,.G are
right and LFLH— LPGC, the triangles are similar, and
GP : PC = FL : FH = AD : FH
= AD: { V (AC. BD) - V(AB . CD)}.
But GP : PG = FP : PB ;
therefore CP 2 : PC 2 = FP . PG : BP . PC