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A history of Greek mathematics - Wilbourhall.org

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THE COLLECTION. BOOKS VI, VII 399<br />

The test <strong>of</strong> apparent equality is <strong>of</strong> course that the two straight<br />

lines should subtend equal angles at F.<br />

The main points in the pro<strong>of</strong> are these.<br />

The plane through<br />

CF, CK is perpendicular to the planes BFE, PFM and LFR<br />

;<br />

hence CF is<br />

perpendicular to BE, QF to PM and HF to LR,<br />

whence BC and CE subtend equal angles at F :<br />

and PQ, QM.<br />

Since FC bisects the angle AFD<br />

t<br />

so do LH, HR,<br />

and AC:CD = AK:KD<br />

(by the polar property), Z CFK is a right angle. And CF is<br />

the intersection <strong>of</strong> two planes at right angles, namely AFK<br />

and BFE, in the former <strong>of</strong> which FK lies; therefore KF is<br />

perpendicular to the plane BFE, and therefore to FN. Since<br />

therefore (by the polar property) LN : NP = IjK : KP, it<br />

follows that the angle LFP is bisected by FN] hence LN, NP<br />

are apparently equal.<br />

Again<br />

LC:CM = LN:NP = LF: FP = LF: FM.<br />

Therefore the angles LFC, CFM are equal, and LC, CM<br />

are apparently equal.<br />

Lastly<br />

LR:PM=LK:KP=LN:NP=LF:FP; therefore<br />

the isosceles triangles FLR, FPM are equiangular; therefore<br />

the angles PFM, LFR, and consequently PFQ, LFH, are<br />

equal. Hence LP, RM will appear to be parallel to AD.<br />

We have, based on this proposition, an easy method <strong>of</strong><br />

a circle<br />

'<br />

solving Pappus's final problem (Prop. 54). Given<br />

ABBE and any point within it, to find outside the plane <strong>of</strong><br />

the circle a point from which the circle will have the appearance<br />

<strong>of</strong> an ellipse with centre C'<br />

We have only to produce the diameter AD through C to the<br />

pole K <strong>of</strong> the chord BE perpendicular to<br />

AD and then, in<br />

the plane through AK perpendicular to the plane <strong>of</strong> the circle,<br />

to describe a semicircle on CK as diameter.<br />

this semicircle satisfies the condition.<br />

Any point F on<br />

Book VII. On the 'Treasury <strong>of</strong> Analysis'.<br />

Book VII is <strong>of</strong> much greater importance, since it gives an<br />

account <strong>of</strong> the books forming what was called the Treasury <strong>of</strong><br />

Analysis (dvaXvouevo? tottos) and, as regards those <strong>of</strong> the books<br />

which are now lost, Pappus's account, with the hints derivable<br />

from the large collection <strong>of</strong> lemmas supplied by him to each

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