UgO METRICAL i»rioi3LEMSd 
38 3 
and upon AF lefc pl parallelogram AFHI be constituted 
equal to the given area of the triangle; make IK per 
pendicular to A I, and equal to FP; and, from the 
point K, to AB, apply KD rz PH; then draw DPE, 
and the thing is done 
DEMONSTRATION- 
Supposing M to be the intersection of CF and IEt, 
it is evident, because of the parallel lines, that all the 
three triangles PHM, PFE* and MDI are equiangular^ 
therefore, all equiangular triangles being in proportion 
as the squares of their homologous sides, and the sum 
of the squares of PF (tK) and DI being equal to the 
square of PH (KD), by construction and Euc- 47. 1. 
it is evident that the sum of the triangles PFE and 
DM! is zz the triangle PHM ; to which equal quan 
tities in Jig. l, let AFPMI be added, so shall ADE be 
likewise equal to AFHI: but, in jig, 2, let PFE be 
taken from PHM, and there will remain EFHM — 
DMI; to which adding AIME, we have AFHI rz: 
ADE, as before. Q. E. D. 
METHOD OF CALCULATION* 
By dividing the given area by the given height of 
the point P above AB, the base Al of the parallelogram 
AFHI will be known, and consequently PH ( = KD); 
whence DI ( rr s/KD 2 ^— PF 1 ) will likewise become 
known.—This problem, it may be observed, becomes 
impossible when KD (PH) is less than KI (PF); which 
can only happen, in case l, when the given area is less 
than a parallelogram under AF and FP. 
PROBLEM LXXI. 
To cut off from a given polygon BCIFGH, a part 
EDBHG equal to a given rectangle KL, by a right-line 
ED passing through a given point P. 
CONSTRUCTION. 
Let the sides of the polygon CB and FG, which the 
dividing line ED falls upon, be produced till they meet 
in A; upon ML [by Euc. 45. 1.) make the rectangle 
c c