GEOMETRY. 
pendicular AB be drawn to that line; and’ different 
tee ue lines AE, AC, AD, &c. to different points of 
that line. - Sue” 
1st," The perpendicular AB is shorter than any 
oblique line. icoes - 
, The two oblique lines AC, AE on opposite sides 
of the perpendicular, and at equal distances BC, BE, 
are equal. 
3d, Of any two oblique lines AC, AD, or AE, AD, 
that which is farther from the perpendicular is the 
greater of the two. ; 
Produce the 
join FC, FD. 
"1. The triangles BCF, BCA are equal (6.), for BF= 
‘BA, and BC is commion, and the angles at B are right 
aes therefore CF = CA, Now AF is less than 
perpendicular, so that BF = AB, and 
+CF, (8.) therefore, taking’ the halves, “AB is less - 
than AC ; that is, the perpendicular is the shortest line 
that can be drawn from A to DE. 
2, Next suppose BE=BC; then the triangles ABB, 
ABC will be eh (6.); for they have also BA com- 
mon, and the angles ABE, ABC equal,’ therefore 
AE=AC; that is two oblique lines ‘equally distant 
from the Rep on opposite sides aré equal. 
3. In the triangle ADF’, the sum of AC and CF is 
less than the sum of AD and DF (9.); therefore AC 
the half of AC+-CF, is less than AD, the half of AD + 
DF’; that iis the oblique line farther from the perpendi- 
een Ts Bieta Shieh POS ahi AN QOAY POLS aie 
Cor. 1, The perpendicular from’a’ 
measures its distance from the line. 
~ Cor. 2. From the same point no more than'two equal 
straight lines can be drawn to terminate in that line. © 
: 
point ‘on a Jine 
19 
Pror. XVII. Turor. 
If through the pote C, the middle of the straight line 
AB, a perpendicular be drawn to'that line. 
1st, Every point in the perpendicular is equally dis- 
Naas, the extremities of the line Py: 
2d, Every point out of the perpendicular, is unequal- 
ly distant from the extremities of the line. re 
1. Because AC=CB, the tw6 oblique lines ‘AD, DB, 
“which are equally distant from the’ endicular, are 
equal (16). The same is also true of the two oblique 
ens mien and of the two ry lines AF. EB, 
&e. erefore, every point in the perpendicular is 
“ye Say opteieee thd andl ofthe line. © 
mt Fd, I be a point out of the perpendicular, . “If TA, 
IB be joined, one of them will cut the perpendicular in 
D; therefore, drawing DB, we have DB=DA. ‘But 
‘TB is less than ID +. DB, and 1D + DB=ID+DA=IA, 
Bart IB a than TA tf ay is, any point out of 
‘the p icular is unequ istant’ from the extre- 
ehities BB. Syrg? a rice | : 
‘Prop. XVIII. Turon, - 
~ "Two righit angled’ triangles are ‘equal, if the hy 
Gina ted Gable ut the coe bs jal to the h path - 
nuse and. a side of the other. be eil Teese 
“Let the’ hyp. 
hyp@henuse ACSDF, and the side AB= 
DE ; the right angled triangle ABC shall be equal to 
the right angled a DEF. 
The proposition will evidently be true, if it can be 
proved that BC=EF (11). Let us suppose, if it be 
“two. lines are. parallel. 
“have AHK + KHC > HKD + KHC; bit the former 
AC, BD, t 
the same side, will be equal to two right angles. 
203 
possible, that these sides are unequal, and that BC is 
the greater. Take’ BG=EF, and join AG. The tri- 
angles ABG, DEF; having AB=DE, and BG=EF 
(by hypothesis), and also having the angle ABG rs 
‘to DEP, they will be equal (6); therefore AG=DF, 
but DF=AC ; therefore AG=AC; that is, two oblique 
lines, one more remote from the perpendicular chit the 
other, are equal, which ‘is impossible (16) ; therefore 
BC is not unequal to EF’: and hence the triangle ABC 
is equal to the triangle DEF, 
Prop. XIX, Tueor. 
If two straight lines AC, BD, are perpendicular to a 
‘third’ AB, these two lines are parallel ; that is, although 
uced ‘ever so far both ways, they will not meef, 
(Def. 11.) 
For if they could meet in a point O, on either side 
of AB, then two perpendiculars OA, OB might be 
drawn from the same point O, to the same straight 
line AB, which is impossible, (15.) 
et XX. _Tueor. 
Tf two straight lines AC, BD, make with a third AB 
two interior angles CAB, ABD, the sum of which ‘is 
equal to two right angles, these two straight lines are 
parallel. 
From G, the middle of AB, draw EGF ndicular 
to AC; then, because the sum ABD + ABF 1s equal to 
‘two right angles (2.),and by hypothesis the sum ABD 4 
‘BAC is equal to two right angles, therefore ABD + 
ABF= ABD + BAC; and taking away the common 
angle ABD, there remains ABF = BAC; that is GBF 
=.GAE. Besides, the angles BGF, AGE are equal 
Aye and BG = GA; wherefore the triangles BGF, 
E have a side and two adjacent angles of the one 
equal to a side and two adjacent angles of the other} 
they are therefore equal (7.), and the angle BFG = 
AEG; but AEG is by. construction a right angle ; 
therefore BFG isa right angle ; and since GEC is aright 
angle, the straight lines EC, FD are perpendicular to 
EF, and are therefore parallel to one another. (19.) 
_. Cor. 1. If two straight lines AC, BD make with a 
third HK the alternate angles AHK, HKD equal, the 
r 
For then, adding KHC, we 
sum is equal to two right angles (2:), therefore the 
latter is equal to'two right angles; and consequently 
AC is parallel to BD. 
Cor. 2. If two straight lines AC, BD are cut by a 
third FG, so as:‘to'make'the exterior angle FHC equal 
to the interior and opposite angle HKD on the same 
side ; the two lines are parallel: For since FHC=AHK 
(4.), then we have AHK = HKD;; that is, the alter- 
nate angles equal, therefore AC is parallel to BD. . 
“ Pror, XX1._ Tuzor, 
Ifa eeraighe line EF meet two parallel ail f lines 
e sum of the inward angles CEF, EFD on 
For if not, suppose EG to be drawn through E, so 
that the sum.of GEF,and EFD may be two right an- 
gles; then EG will be parallel to BD (20.) ; and thus 
through the same point E two straight lines EG, EC 
Principles. 
—_—— 
Fig, 41. 
Fig. 42. 
Fig. 43, 
Fig. 44. 
