Principles. 
—_—\—_ 
Fig. 43, 
Pig. 44, 
Fig, 45, 
Fig. 46. 
204 
are drawn, each parallel to BD, which, is impossible, 
(Ax. 11.); therefore no straight line that does not.co- 
incide with AC is parallel to BD, therefore the straight 
line AC is parallel to BD. : val} 
Cor. 1. If a straight line is perpendicular to one,of 
‘two parallel straight lines, it is also perpendicular to the 
other. 
Cor. 2. (Fig. 43.) If a straight line HK meet two 
parallel straight lines AC, BD, the alternate angles 
AHK, HKD shall be equal. Sy 
For the sum CHK4HKD is equal to two right an- 
gles; and the sum CHK-+AHK is also equal to two 
right angles (2.), therefore the angle HKD must be 
equal to AHK. 
Cor. 3. (Fig. 43.) If a straight line FG cut two, pa- 
rallel straight lines AC, BD, the exterior angle FHC 
is equal to the interior and opposite angles HKD. 
For since FHC=AHK (4.), and AHK = HKD; 
therefore FHC = HKD. 
Scuorium. If a straight line EF meet two other 
straight lines EG, FD, and make the two interior an- 
gles EFD, FEG on the\same side less than two right 
angles, the lines EG, FD meet, if produced, on the 
side of EF, on which the angles are less than two right 
angles. For if they do not meet on that side, they are 
either parallel, or else they meet on the other side. 
Now they cannot be parallel, for then the two interior 
angles would be equal to two right angles, instead of 
being less, Again, to shew that they cannot meet. on 
the other side, suppose EA to be parallel to DFB ; then 
because the sum EFD +.FEG is (by hypothesis) less 
than two right angles, that is, less than FEK + FEG 
(2.); and EFD = FEA (Cor. 2. of this Prop.) ; there- 
fore the sum FEA '+ FEG is less than FEK + FEG; 
and, taking FEG from both, FEA is less, than FEK ; 
hence FB and EK must be on opposite sides of EA, 
and therefore can never meet: 
The truth of this proposition is assumed as an axiom 
in the Elements of Euclid, and made the foundation of 
the theory of parallel lines, 
Prop. XXII. Turor. 
- Two straight lines AB, CD, parallel to a third EF, 
are parallel to one another, 
Draw the ‘straight line PQR perpendicular to EF. 
Because AB is parallel to EF, theling PR shall be per- 
pendicular to AB (1. Cor. 21.) And because CD a - 
rallel to EF, the line PR. is also perpendicular to.CD ; 
therefore AB and CD are perpendicular to the same 
straight line PQ ; hence they are parallel. (19.) 
Prop. XXIII. Turor. 
Two parallel straight lines are every where equall 
distant” ig ry qually 
Let AB, CD be two parallel straight lines. From 
any points E and F in one of them, suppose perpendi- 
culars EG, FH to be drawn ; these, when preduced, will 
meet the others at right angles in H and G (1. Cor. 21.): 
Join FG; then, because FH and EG are both per- 
pendicular to AB, they are parallel, (19.); therefore 
the alternate angles HFG, FGE which they make with 
FG, are equal, (2. Cor. 21.): And because AB is paral- 
lel to CD, the alternate angles GFE, FGH are also 
equal ; therefore the triangles GEF, FHG have two an- 
‘to each ; and the side FG adjacent to the equal a 
common ; the sn ff aN ry 
GEOMETRY 
gles of the one equal to two angles of the 
68h 
FH is equal to EG, that is, any two points . on one 
of the lines, are equally distant from the other line. 
So) htPRoR: KKIV. «: TagoRe: ee. 
In any triangle, if one of the sides be produced, the 
exterior ae ual to both the interior and opposite 
angles ; and the three interior angles are equal to two 
right angles, - 
Let ABC be a triangle; produce any one of its sid 
AC towards D, and from the point A,let AE fog mig ~ 
parallel to BC: And because of the peels Ae 
AE, the angle. EAD =C, and the angle EAB= B. 
(Cor. 2. and 3. of Prop. 21.95 therefore EAD + EAB= 
C +B; that is, BAD=B4C: Hence the outward 
angle is equal to the two inward and opposite an- 
les. a pay JEN Ley 
Again, because BAD = B +, to each, add BAC, 
and we have BAD + BAC=B+C BAC, that is, 
equal to the sum of the three angles af the triangle ; 
but the sum BAD + CAB is equal to two right ane 
(2.); therefore the sum of the three angles of the tri- 
angle ABC is equal to two right angles. = ss 
Cor: 1. If two angles of one triangle are e 
two angles of another triangle, each to each, the thir 
angle‘ of the one shall be equal to the rep of the 
other, and the triangles shall be e uiangular. 
Cor. 2. A triangle can have only one right angle. 
Cor.,3. In any right angled MRE gs sum of the 
two acute angles is Fava toaright angle. 
Cor. 4. In an equilateral triangle, each of the angles 
is one-third of two right angles, or two-thirds of one 
right angle. 9 ' 
ienta sit to oll i+.) Stusdct 3 cipro VE 
3 Pror, XXV. THEoR. hs 
The sum of all the interior angie of a polygon is 
equal to twice as many right angles, wanting four, as 
the polygon has sides. 
Let ABCDE, &c. be a polygon ; if from the vertex of Fig: 
any one of its angles A, diagonals AC, AD, AE, &c, 
be drawn to the vertices of all the o angles, it is 
evident that, the polygon will be divided into five tri- 
angles if it have seven sides ; and into six triangles if 
it fave eight sides ; and, in general, the number of tri 
angles will be two less than the number of sides. It is 
also evident, that the sum of all the angles of these tri- 
angles make up all the angles of the polygon ; now, all 
the angles of each triangle are feast er equal to two 
right angles ; therefore,all the angles of. the triangles, 
that is all the angles of the polygon, will be equal to 
two right angles taken as often, except two, as e fi- 
gure has sides, and consequently all the angles of the 
polygon will be equal to twice as many right angles 
wanting four as the figure has sides. 
Cor. The sum of the angles of a quadrilateral will 
be four right angles. sien 
Scuorium. The proposition will only apply to such 
polygons as have their angles salient, that is when the 
straight line that joins. any two adjacent angles falls 
within the polygon. When some of the angles are 
saa e proposition must have a different t 
orm, 
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