This Euclidean definition of the tangent is pivotal to Gregory's contradiction. For if KC is not tangent to OEK, it means that some point a on KC must lie above the curve OEK. In particular, if DF is the line segment perpendicular to OA which intersects OA at D, OEK at E, KC at a, and OFL at F, then DE < Da by choice of a. This is shown in the figure below (adapted from Gregory's proofsee the Appendix ).
Actually, the figure tells only half of the story. It's also possible that a could lie on the other side of K on the line CK. However, the argument for that case is similar, and the details make a good exercise (or see the Appendix ).
By the definition of the curve OEK, if area(OFD) denotes the area enclosed by the curve OF and the line segments OD and DF, then DE = area(OFD). Likewise, IK = area(OFLI). Thus, IK/DE = area(OFLI)/area(OFD). Since DE < Da, it follows that IK/Da < IK/DE and so

IK
Da

< 
area(OFLI)
area(OFD)

. 

Notice that
C,
a, and
K are collinear and that
IK and
Da are both perpendicular to
OA. Thus, the triangles
CDa and
CIK are similar. So by corresponding sides,
IK/
Da =
IC/
DC and hence
IC/
DC <
area(
OFLI)/
area(
OFD). Now multiply the numerator and denominator in
IC/
DC by
IL to obtain
IC/
DC =
IC·
IL/
DC·
IL. It follows that

IC·IL
DC·IL

< 
area(OFLI)
area(OFD)



This last step may seem mysterious, but it serves the purpose of converting the original inequality into an inequality about areas which will lead Gregory to an obvious geometrical contradiction. Specifically, DC·IL is the area of the rectangle with sides IL and DC, and IL·IC is the area of the rectangle with sides IL and IC. Inverting the ratios in the last inequality yields:

area(OFD)
area(OFLI)

< 
DC·IL
IC·IL



Now we can make some observations about these areas to arrive at a contradiction: Note that
area(
OFLI) =
area(
OFD) +
area(
DFLI) and likewise, since
IC =
DI +
DC,
IC·
IL =
DI·
IL +
DC ·
IL. Substituting these expressions into the numerators of the last inequality and doing some algebra, we have


area(OFLI)  area(DFLI)
area(OFLI)

< 
IC ·IL  DI·IL
IC·IL





Û 1  
area(DFLI)
area(OFLI)

< 1  
DI·IL
IC·IL





Û 
DI ·IL
IC·IL

< 
area(DFLI)
area(OFLI)

. 


But recall the defining equality for Cnamely, IK/IC = IL. This implies that IK = IC·IL. Since IK = area(OFLI) by definition of the curve OEK, the denominators in the last inequality are the same. Therefore, the numerators must satisfy DI·IL < area(DFLI). But remember that OFL is increasing by assumption. Thus, the rectangle with sides IL and DI must circumscribe the region DFLI.
Hence area(DFLI) < DI ·IL as well. This contradiction shows that KC must in fact be tangent to OEK. Hence the fundamental theorem of calculus follows.