348 RELATION BETWEEN THE AREA OF A PLANE 
the same straight line. Cut off CE equal to CB, and CK equal to 
CA; and jom EK. Then (4.1. E) the triangle ECK is every way 
equal to the triangle ABC. Therefore triangle ECK=triangle FCD ; 
and consequently triangle EDH=triangle FKH. Cut off HM equal 
HE, and HP equal to HD; and join MP. Then triangle HMP is 
every way equal to triangle EDH. Therefore triangle HMP=tri- 
angle HKF; and consequently triangle KMN=triangle FNP. The 
point P cannot fall beyond F, so as to make HP greater than HF ; 
for, if it did, the poimt M would (in order that the triangle HKF 
may not be a part of the triangle HMP) fall between K and H; in 
which case the angle F would be greater than the angle HPM ; that is, 
F would be greater than the angle HDE; whereas, since the exterior 
angle of a triangle is greater than either of the interior and opposite 
angles, the angle HDE is greater than F. In like manner it can be 
proved that the point P does not coimcide with F. And therefore P 
is between H and F; which implies that M is beyond K im the le 
HKM. Hence, from the two given equal triangles ACB and FCD,. 
with the angles at C equal to one another, we have passed to the equal 
triangles KMN and FNP, with the angles at N equal to one another. 
Let 8, be the sum of the angles of the triangle KMN ; and s, the 
sum of the angles of the triangle FNP. Then 
S,—s,= M+ MKN—(F + FPN) 
= E+ EKC —(F + FDC) 
eee —(F + FDC) 
=A irs) a 
Let the same construction that was made with reference to the tri- 
angles ABC and FDC be now made with reference to the triangles. 
KMN and FNP; that is to say, cut off NQ equal to NM, and Nx 
equalto NK. Jom Qr. Cut off RL equal to RP, and RT equal to 
RQ. Join TL, cutting NF inh. Then Q must lie beyond P, on the 
Ime NPQ; for, if it did not, the pomt 7 would lie beyond F on the 
lme NrF; in which case the angle Q would be greater than the angle 
NPF; that is, the angle E would be greater than the angle CDF: 
which is not true. And the point Q lying beyond P, the point r 
must fall between N and F. Hence, as above, we can prove that the 
triangles Trk and FLA are equal to one another; and, if 8, be the 
sum of the angles of the triangle Tvh, and s, the snm of the angles 
of the triangle FLA, 
S,— s,=8,-—5,;=S8—-s. 
