430 
101 
GEOMETRY. 
In the ifofcdii or tquilakral triangle AC B ; a line 
0 drawn from the vertex C to the middle 
of the- bafe A B h perpendicular to the 
iafe, and bifeEls the vertical angle : 
and the contrary; that is, if a line 
drawn perpendicular to the bafe 
of a triangle bifedts the vertical 
angle ; then that triangle mud be 
ifofceles, and the perpendicular is 
drawn from the middle of the 
® bafe.— The triangles ADC, BDC, 
are congruous; fince CA=CB (23); CDrriCD, and 
AD=DB by fuppofition ; therefore ^ A=^B, ^ACD 
“^BCD, 2:^ADC=BDC (ioi); conlequently CD is at 
riglit angles to AB, 
104.. Carol. Hence in any right-lined triangle where 
there are equal fides, or angles ; the angles A, B, op- 
pofite to equal fides B C, AC, are^equal ; and the fides 
BC, AC, oppolite to equal angles, A,*^, are equal. 
In every right-lined triangle ABC, the greater angle 
C is appofite to the greater ftde A B, 
—In the greater fide AB, take 
A D=A C ; draw C D, and througli 
B draw B E parallel to C D ; then 
the angles A D C, A C D, are equal, 
(104); and ^ADC=:^ABE; 
therefore ^ACD = ^ABE ; 
that is, a part of the angle ACB 
is greater than the angle A B C : 
confequeptly the /fZ is greater 
in the fame manner it may be 
A D E 
than the ^ B ; and 
proved to be greater than the ^A, if the fide AB be 
greater than C B. 
106. Carol. Hence in every right-lined triangle, the 
greater fide is oppofite to the greater angle. 
107. Parallelograms A C D B, E C D F, EG H F, Jlattd^ 
ing on the fame bafe CD, or on equal bajes CD, G H, and 
between the fame parallels, C H, A F, 
o.re equal. —For A Bz=E F, being 
each equal to C D ; to each add 
B E, and A E—B F ; now AC= 
B D, and CE=:DF; the trian¬ 
gles A C E, B D F, are therefore 
congruous, (lOi). Now if from 
each of the triangles, ACE and 
-_- - -^ BD F, be taken the triangle 
vy -U G ±1 B IE, the remaining trapeziums 
ABIC and F E 1 D, are equal : then if to each of the 
trapeziums ABIC, F E I D, be added the triangle 
C 1 p, their fum will be the parallelograms AD and 
D E, which are equal. And in like manner it may be 
fiiewn, that the parallelogram EH is equal to the pa¬ 
rallelogram ED=AD. 
jO?. a triangle ABC is the half of a parallelogram 
A D, when they fand on the fame 
bafe A B, and are between the fame 
parallels AB, C D.—A C is e. 
qual to D B, and A B to D C ; 
alfo B C is a fide common to 
both the triangles ABC and 
D C B : thefe triangles are there¬ 
fore congruous, (101.) Confe- 
quently the triangle ABC is 
half the parallelogram AD, 
109, Corot. 1. Hence every parallelogram is bifcfled 
by its diagonal. 
110. Carol. 3. Alfo, triangles Handing on the fame 
bafe, or on equal bales, and between the fame parallels, 
are equal; tliey being the halves of equal narallelo- 
grtyns under like circumllances. 
111. In every right-angled triangle B A C, the fquart emt 
the fide B C oppofte to the right angle A is equal to the fum 
of the fquarts of the two fdes A B, 
AC, containing the right angle. —On 
the fides AB, AC, B C, con- 
flruft the fquares AG, A E, CD, 
( 68 ;) draw AD, CE; and draw 
AF parallel to BD; then the 
triangles A B D, E B C, are con¬ 
gruous. For the ^ A B E = 
C B D, being right angles ; to 
each add the angle ABC, then 
^ E B C and ^ A B D are equal; 
therefore E B, B C, ^E B C, are refpeflively equal ta 
AB, B D, ^ABD. Alfo the triangle E B C is Iralf 
tlie parallelogram A E, (108 ;) for they Hand upon the 
fame bafe EB, and are between the lame parallels 
EB and AC; BA making right angles with BE and 
CA continued. Likewife the triangle ABD is half 
the parallelogram B F ; for they Hand upon the fame 
bafe B D, and are between tlie fame parallels B D, 
A F ; therefore, as the halves of the parallelograms- 
]'A and BF are equal, confequently the parallelogram 
B F is equal to the fquare AE. 
In the fame manner may it be flievvn, that the paral¬ 
lelogram C F is equal to the fquare AG : but the pa¬ 
rallelograms B F and C F together, make the fquare 
C D ; tlierefore, the fquare C D is equal to the fquares 
E A and A G. 
112. Corot. 1. Hence if any two fides of a right angled 
triangle are known, the other fide is alfo known.— 
For B C = fquare root of the fum of the fquares of A C 
and A B. 
AC = fquare roo-t of the difference of the fquares- 
of B C and A B. 
A>B = fquare root of the difference of the fquares 
of B C and A C. 
1 13. Or thus, making the quantities BC*, AB , AC , 
to Hand for the fquares made on thofe lines. And the 
mark 1/ to Hand for the fquare root of fuch quantities 
as Hand under the line joined to the top of this mark. 
Then B C = y'A C^-f- A B'*; 
AC=\/ B C*—AB^ 
AB = v' —AC } 
Scholium. The lines of the lengths 5, 4, 3, (or their 
doubles, triples, &c.) will form a right-angled triangle- 
For 32 —oj- 25 = 16-1-9. 
114. Corel. 2. Of all the lines drawn from a given 
point to a given line, the perpendicular is the fiiortefl. 
115. Cerol. 3. The fhorteH difiance between two pa¬ 
rallel right lines, is a right line drawn from one to the 
other perpendicular to both. 
116. Coro/. 4. Parallel right lines are equidiHant; 
and the contrary : for two oppofite fides of a rectangu¬ 
lar parallelogram are equal (2S) ; and each is the Ihort- 
elt diHance between the other fides, 
1x7. If a right line A'R, be divided into any two parts 
AC, C B ; then will the fquare on the 
whole line be equal to the fum of the 
fquares on the parts, together with two 
reblangles under the two parts. —That is, 
AB^^ac"* -fUB^ -f-3XACxCB. 
The rectangle under two lines is ge¬ 
nerally expreffed by three letters; 
the firH two letters*and for one line, 
and the laH two for ?he other line, Thu s for AC XC B,. 
is written A C B : for ABxBC, is w iitten A B C ; and. 
c 
T 
B 
G 
E 
ior aXACxCB, is written aACB.—Let AD, AF 
by' 
