I i6l ] 
P R O B L E M. 
The Line being divided any where in^, todmie 
k again in x between ^ and c fo that the Redangle ax^ 
(hall be equal to the Rectangle ^J^^ together with the 
double fquare of xc. . 
r ^ } ' — I 4- 
h X c d 
ANAL Y S I S. 
Let therefore ax(f = bxc 2xC3t 
But by 3. 2. EL bcx = bxc h- xcx 
Therefore axb == bcx xcx 
Let be made ^ h , theref. bcx = dcx 
Therefore axb = dcx = xcx 
that is by 3. 3. EL . axb = dxc 
Therefore ax, xc : : xd^ bx 
zviA Componendo ax^ xc : : db, bx 
Let cf hQ made ±= bd cf 
andas the fum of the Antecedents , to the fum of the 
Gonfequents. So one Antecedent to its Confequent* 
Therefore af, be : : cf, bx» 
Therefore the Problem is folv'd, 
ConfiruB'ton and Demon fir at ion* 
Let cd and Af be made equal to bc^ and let af^ be, cf, h^^t 
be proportional , I fay the thing is done* 
For fince af, hoix r/, hx^ and the difference of 
the Antecedents to the difference of the Confe- 
qcuences as one Antecedent is to its Confequenc^^r will be 
to as cf or he to hx , and the Reftangle axb will 
be equal to the Reftangle ixc^ that is, to the Reft^n- 
gle i^-x together with the Square of xc or (becaufe h& 
and cd are equal ) to the Redangle bcx with the Square 
of xr 5 But the Kedangle^^.v is equal to the Rediogle 
I'xc and the Square of xc : Therefore the Rt dangle Axb 
is equal to the Redangle ^^r^ and the double Square of 
^•s^. Which was to be done. 
The foUomng Profojimn is taken out of the 4th# Bo&k^ 
P R O F:^L E Mo 
T WO PDii3ts a and! i' being gi vefl j to dfaw the two 
Lmei . 
