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