THE COLLECTION. BOOKS VI, VII 
399 
The test o£ apparent equality is of course that the two straight 
lines should subtend equal angles at F. 
The main points in the proof are these. The plane through 
CF, CK is perpendicular to the planes BFE, PFM and LFR; 
hence CF is perpendicular to BE, QF to PM and HF to LB, 
whence BG and GE subtend equal angles at F: so do LH, HR, 
and PQ, QM. 
Since FG bisects the angle AFP, and AG: CD = ZL7T: KD 
(by the polar property), Z CFK is a right angle. And CF is 
the intersection of two planes at right angles, namely AFK 
and BFE, in the former of which FK lies; therefore KF is 
perpendicular to the plane BFE, and therefore to FN. Since 
therefore (by the polar property) LN: NP — IjK : KP, it 
follows that the angle IjFP is bisected by FN; hence LN, NP 
are apparently equal. 
Again LC:CM = LN:NP = LF:FP = LF:FM. 
Therefore the angles LFG, GFM are equal, and LG, CM 
are apparently equal. 
Lastly LR: PM=LK: KP=LN: NP=LF:FP; therefore 
the isosceles triangles FLR, FPM are equiangular; there 
fore the angles PFM, LFR, and consequently PFQ, LFH, are 
equal. Hence LP, RM will appear to be parallel to AD. 
We have, based on this proposition, an easy method of 
solving Pappus’s final problem (Prop. 54). ‘ Given a circle 
ABDE and any point within it, to find outside the plane of 
the circle a point from which the circle will have the appear 
ance of an ellipse with centre G.’ 
We have only to produce the diameter AD through G to the 
pole K of the chord BE perpendicular to AD and then, in 
the plane through AK perpendicular to the plane of the circle, 
to describe a semicircle on CK as diameter. Any point F on 
this semicircle satisfies the condition. 
Book VII. On the ‘Treasury of Analysis’. 
Book VII is of much greater importance, since it gives an 
account of the books forming what was called the Treasury of 
Analysis (draXvoyeros totto?) and, as regards those of the books 
which are now lost, Pappus’s account, with the hints derivable 
from the large collection of lemmas supplied by him to each