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PROBLEM. 
The Line rff being Divided in ^ to Divide it again 
in X between ^ and b fo that ^jc, ^c^ be Proportional. 
Now becaufe in the Proportion ax^occ : : cx^ xh^ the firft 
Ratio is Acidi:ive and the fecond Suhtra^ive it is evident 
that the Additive muft either be made StibttABive^ or the 
suhtYAiiive Additive. But becaufe the Terms are un- 
known, let^c be bifeded in my aod 2 m x will be 
the Difference of the Parts ax^ xc ; likewife let k be 
bifeded in p , and 2xp. will be the fum of the Parts xc 
and xb 5 whence one may proceed by Compofition or 
Di/Vifion. 
A/tafyJif. 
Let 4X, xc : : xc, xi 
Theref. Componendo ac xc:: 7xp^ xh 
and half^ the Antecedents ^ ntc^ xc : : xp, xb 
and Comertende mc mx : : xpy bp 
Therefore the Problem is folv'd^ Becaufe the Point % 
being only in the middle Terms, we can proceed no far- 
ther. And becaufe there is nothing from whence we may 
infer which of the two mx and xp is the greateft, it will 
be in our choice to take w^iT either for the greateftorthe 
lead part, and there will be two Solutions for 
which there is one Demonftration. 
Ccnftruiiion and Demonftration. 
Let ac be bifeded in m and be in p^ andto mc and bp 
QMpcl^t two Reciprocals andx/'befound whofe fum 
bcmp^ I fay the thing is done. 
For by the ConftruLlion wr, mx : : xp, ip y There- 
fore Convertendo mcy xc : : xp^ xb and doubling the 
Antecedents ac^ xc: : 2xp, xby but txp is the fum 
of 
