346 RELATION BETWEEN THE AREA OF A PLANE 
FV, VQ,......, SN; the number z being taken so great that ” times 
MN is greater than the given line AB. Let TV, PQ, &c., be per- 
pendicular to NF. Suppose then the base DC of the triangle ADC 
(Fig. 6) to diminish, according to the law of its variation, until CD 
becomes less than FV ; and, if AC be not less than AD, produce AD 
to K, making AK=AC. Join CK; draw AR perpendicular to CK ; 
and cut off the parts Rz, zy, ya, &c., each equal to MN, until AR is 
exhausted ; the last part being possibly less than MN. At the points 
of section, z, y, x, &e., raise the perpendiculars zw, yh, vb, &c. Then, 
because CD is (by hypothesis) less than FV or NS (Figs. 5 and 6), 
and because it is obviously greater than CR, NS is greater than CR. 
Also, because times MN is greater than AB, and AC is (by hypo- 
thesis) not greater than AB, 2 times MN is greater than AC. Much 
more is 7 times MN greater than AR. And the parts Rz, zy, &c., 
were cut off each equal to MN. Hence the number of such parts is 
not greater than 2; and the number of the spaces, 
RUD aTiy” YDS CO OPIN MERE: Se REN ae es ce 
into which the triangle ARC is divided, is not greater than the num- 
ber of the spaces, 
BEDS) NADAS oy ole ge UNIVE ie i o's ee 
into which the figure MNFH has been divided. But since NS is 
greater (as we have proved) than RC, and MN is equal to Rz, the 
space RzwC may be wholly inserted within the space MNS¢, and is 
therefore less than that space. But RzwC is the greatest space in 
the series (1), and MNS¢ is the least in the series (2). Hence, since 
the number of terms in (1) is not greater than the number of terms 
in (2), the sum of the terms in (2) is greater than that of the terms 
in (1): that is, the triangle ACR is less than MNFH. Hence the 
triangle AKU is less than the circle EHG. Much more is the triangle 
ADC less than the space L. 
In the next place, suppose, if possible, that, as CD is indefinitely 
diminished, the sum of the angles of the triangle ACD ultimately 
differs from (in which case it must, by Prop. I, be less than) two 
right angles by more than the finite angle BAH (Fig. 7); BA being, 
as in the previous case, a given line which neither of the sides, AC, 
AD, ever exceeds. Produce HA to any point W, and AB to any 
point E. Join EW; and draw BV perpendicular on EW. Let the 
base DC (Figs. 6 and 7) be diminished, according to the law of its 
