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A history of Greek mathematics - Wilbourhall.org

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THE COLLECTION. BOOK IV 375<br />

Therefore (Lemma I), it' the two circles touch the semicircle<br />

BED in R, E respectively, FRE is a straight line and<br />

EF.FR = FH\<br />

But EF.FR = FB 2 ; therefore FH = FB.<br />

and MA produced in S, we have,<br />

If now BH meets PN in<br />

by similar triangles, FE:FB = PH:PO = AH: AS, whence<br />

PH = PO and SA = AH, so that 0, S are the intersections<br />

<strong>of</strong> PN, AM with the respective circles.<br />

Join BP, and produce it to meet MA in K.<br />

Now<br />

BM: BN=FA: FP<br />

And<br />

= AH-.PH, from above,<br />

= AS:PO.<br />

BM:BN=BK:BP<br />

= KS : PO.<br />

Therefore KS — AS, and KA =

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