216 
"Proportion to the'sum of the squares on the two sides containing 
eae the right angle. yd: 
Fig. 100. 
fig. 101. 
Fig. 102. 
» ducing it to E ; 
:the trian 
Let ABC be right angled triangle, of which A is 
~ the right angle. . Having formed, squares on the three 
sides, draw AD perpendicular to the hypothenuse, pro- 
raw also the likes AFOH. 
The angle ABF is made up of the angle ABC and.a 
right angle CBF ; the angle: HBC is made up of the 
-same angle ABC and a ri: 
vangle ABF = HBC: but AB = BH, because they are 
angle ABH ; therefore the 
sides of the square, and BF = BC for a like reason ; 
therefore the triangles ABF, HBC are equal (6. 1.) Now 
ABF is half of the rectangle BDEF, or BE, 
because they have the same base BF and the same alti- 
pri! 3 and the triangle HBC is in like manner 
half of the square BL, for they have the same base BH, 
and the same altitude; for, because the angles BAC, 
BAL are right angles, the lines CA, AL form a conti- 
nued straight line (3: 1.), which is parallel to BH ; 
therefore the rectangle BE is equivalent to the square 
BL. In like manner, by joining AG and BI, it may 
be demonstrated that the rectangle .CE is equivalent to 
the square CK; therefore the two rectangles BE, CE 
are together equal to the two squares ‘BL, CK: but 
these rectangles make up the whole square on BC, ‘the 
side opposite to the right angle, and BL, CK are the 
squares on BA, and. AC, the \sides containing the right 
angle ; therefore the square on the side subtending the 
vigpt angle is equal to the sum. of the squares-on the 
sides containing the right angle. 
Prop. XII. Tneor. 
In any triangle ABC, the square of AB, the side op- 
-posite to any one of sits acute ~angles, is less than) the 
sum of the squares of the sides AC, CB, which contain 
that angle ; and if a perpendicular. AD be drawn to 
either of these BC from the opposite angle, the differ- 
ence shall be equal to twice the rectangle BC x CD, so 
that , 
AB?= AC? + CB? -—~ 2 BC x CD. 
There are two eases, according as the perpendicular 
falls within or without the triangle. In case first, BD= 
BC—CD>; and in case second; BD=DC— BC. In 
either case, BD? = BC?+CD? —2 BC x CD (9. To 
each of these'equals add DA?, and we have 
BD* + DA*= BC? +CD? + DA2—2BC xCD; 
but BD? 4 DA?=BA?(11.), and CD?4+-DA:=CA?; 
therefore BA°=BC?-+4 CA?—2 BC x CD: : 
Prop. XIII. 
In any‘obtuse angled triangle ABC, the square of 
AB, the side opposite to the obtuse angle, is greater 
than the squares 6f AC, BC, the sides containing the 
obtuse angle ; and if a perpendicular AD be drawn on 
either of these sides, the excess will be equal to 2 BC x 
CD ; so that we have ~ 
AB*=AC* 4+ CB:4.2BCxCD. 
THEOR. 
For BD=BC+CD ; therefore BD?= BO? 4. CD? 
2BCxCD (8.); to each of these equals ada DAL 
and we have 
‘BD? + DA?=BC?4 CD24. DA? +2BCxCD. 
But Bb? 4+ DA?= BA?; and CD?.4. DA?=CA?; 
therefore BA* =.BC* 4+. CA? +2BC x CD, 
GEOMETRY. | 
the 
valent, (2. 
altitude, they are to each other as their 
ens F 4 Cheng iy 
Prop. XIV. Taror. — ult cesip. 
eh ok 7 > Ooh s Sa lee 
In-any triangle ABC, if a-straight lime AE bedrawn ,;, 
from is vertex t the mile of the base, the sum of "ig 
squares of the sides is equal to. twice ‘the square of 
that line, and twice the square of half the base. velar tt 
ty ANS dds Be ole 
res j mt 
Draw AD perpendicular to the base: Then’ / 
AB:= AE? + EB? 4+2BE xX ED (13.) 0° ~ 
AC? = AE? 4+ EC*—2CE x DE (12.) °. 
Hence, by adding, and observing that BE and. 
therefore BE? = CE?, and BEx ED=CEXED, we 
get AB? 4 AC?=2AE?42BE% 8 0h 
P ; 
ohova 
‘Page. Xv. “‘Tugon, * vt oH 
A straight line DE drawn parallel to the base of a pj, 
triangle ABC, divides the sides AB, AC proportionally; 
or so that AD: DB:: AE: EC, ) <) eo ede 
. ; 10) & : L Ts 
Join BE and DC; the two triangles BDE, CDE have 
the same base DE; they have also the same altitude, 
because BC is parallel to DE; therefore _ are equi- 
Again, because the triangles ADE, E 
have manifestly the same altitude, they are to one ano- 
ther as their s, that is ADE: BDE:: "AD: BD. 
Also because the triangles ADE, CDE have'the same 
ADE:CDE::AE:CE; but we have seen they se “a 
triangle BDE = triangle CDE; therefore, because of 
the common, ratio.in the two proportions, it follows 
that AD’: BD;:: AE: CE. 
Prop. XVI. Tueor. 
Conversely, if the sides AB, AC be cut proportion- p 
ally by the line DE, so that AD: DB:: AE: EC, the 
line DE shall be parallel to the base BC, mae 
¢ 
For if DE be not parallel to BC, some other line DO 
will be parallel to BC: then, by the preceding theorem, — 
AD:DB:: AO: OC, but by hypothesis, AD: DBr: 
AE: EC; ‘therefore, AO: OC :: AE: EC; and b: 
composition, AC: OC: : AC: EC; hence OC must 
equal to EC, which is impossible, unless.the point O 
fall at. E; therefore no line besides DE can be parallel 
to BC. Hy vise? ee “Gitpo 
. 
sy 
Prop. XVH. Tueor.. zi, 
If a straight line BD be drawn from the vertex of a 5. 
triangle, so as to make equal angles:with ‘its sides BA, ~' 
BC, the: distances of the point D, an which dt eutsithe | 
base from A and.C, the extremities of the base, shall — 
have to each other the same ratio as the adjacent sides - 
BA, BC-of the triangle: that is AD: DC::AB?BC, — 
: + aie o otrot jilles 
From C, one extremity of the base, draw CE)paral- — 
lel to BD, meeting AB in-E. ‘The angle-ABD=BEC 
and the angle CBD=BCE.(2 and 8 Cor. 21. 1,), but 
by hypothesis the angle ABD=CBD, therefore'the — 
angle BEC=BCE ; hence the side BC = side BE, (1%.) _ 
Again, because A BD is:a triangle, and CE is: drawn 
parallel to.one of its sides, AD »DC::AB: BE; but — 
it has been shewn that BE=BC; therefore AD: DC ;: 
AB: BC.. ' OL 4A sreraehy 8 ae 
‘Schotium There may be tivo cases, in’ one the line 
i) 
