344 RELATION BETWEEN THE AREA OF A PLANE 
Draw F'N perpendicular to CF, and equal to HD. Produce it to L, 
making LN = HD; and jom HL and HN. Then the sum of the 
angles of the triangle CHD is not less (Cor. 1. Prop. I.) than S; 
that is, it is not less than two right angles. Therefore (Prop. 1) it is 
equal to two right angles. But (4.1. E.) the triangles CHD and 
FHD are every way equal. Therefore angle HCD = angle HFD. 
But the sum of the angles DCH and DHC has been proved to be 
equal to aright angle. Therefore the angle CHD = the angle DHF 
= the angle HFN. Therefore (4.1. E.) the triangles DHF and 
HFN are every way equal; and hence HNF is a right angle. Conse- 
quently (4. I. E.) the triangles HNF and HNL are every way equal. 
Hence 
ZLHF + 2CHF = 2 2NHF + 2 2CHD 
2 ZHFD + 2 2CHD 
2 ZHCD + 2 zCHD 
= 2 right angles. 
Therefore CHL is a straight line. Also the sum of the angles of the 
triangle LCF is equal to two right angles. Hence, beginning with 
the hypothesis that the sum of the angles of the triangle CHD is 
equal to two right angles, we have found that the sum of the angles. 
of the triangle LCF is equal to two right angles; the sides of the 
latter triangle being double those of the former. By going on in the 
same manner, we can find a triangle ABC, with one of its angles BAC 
a right angle, and the sum of all its angles equal to two right angles ; 
and having each of the sides greater than any given line. Suppose 
now that zy z (Fig. 4) is any triangle what- 
soever ; & y being not less than either of the 
other sides: in which case, as DH (Fig. 3) z 
falls within the base CE of the triangle FIG 4,—]~ 
HCE, the perpendicular z ¢ from 2 (Fig. 4) — 
upon zy falls within the line # y. Then x % 
= 
the triangle BAC (Fig. 3) being constructed 
in the manner above described, so that each ‘ 
of the sides BA and AC may be greater than any of the lines wz, ay, 
yz, in Fig, 4, cut off MA equal to z¢, and AP to zé The sum of 
the angles of the triangle BAC is not greater (Cor. 2, Prop. I.) than 
the sum of the angles of the triangle PAM or zzt. That is, the gum. 
of the angles of the. triangle wz¢ is not less than two, right angles. 
Hence (Prop. I.) it is equal to two right angles. In like manner the 
