Encyclopaedia Britannica; or, A dictionary of arts and sciences, compiled upon a new plan … > Volume 2, C-L

<G TE 'G m
>64. IF from ihe centre fC of a circle ABE, there be
-ctlrawn a perpendiciilar CD on the chord AB, and
produced till it meet the circle in E, then the line
‘CF bifedts the arch AB in the point F ; for (fee the fbre-
:going figure) joining the points A and F, F and B by the
llreight lines AF, -FB, then in the triangles ADF, B.DF,
AD is equal to DB (by art. 63.) and DF common to
Both ; therefore AD and DF, two legs; of the triangle
ADF, are equal to BD and DF, two-legs of the trian¬
gle BDF, and the included angles ADF JBDF are equal,
being both right ; therefore (by art. 61) the remaining
legs AF and FB are equal; but in the feme circle equal
lines are chords of equal arches, therefore the arches'AF
and FB are equal. So the whole arch AFB is bifedted^
in the point F by the line CF.
Cor. 1. From art. 63. it follows, tjiat any line bifedt
ing a chord at right angles is a diameter; for fi nee .(by
art. 63.) a line draw from the centre perpendicular to *
chord, bifedts that chord at right angles ; therefor*, con-
-verfly, a line bifedting a chord at right angles, muft pals
thro’ the centre, andeonfequently be a dii.meter
Cor. 2. From the two lad articles it follows, that-the fine
of any arc is the half of the chord of twice the arc; (or
•(fee the foregoing fcheme) AD is the fine of the arc
AF, by the definition of a fine, and AF is half the arc
AFB, and AD half the chord AB (by art. 63.); there¬
fore the cor. is plain.
65. In any triangle, the half of each fide is the fine
of the oppofite angle ; for if a circle be fuppofed to be
drawn through the three angular points A, B, and D of
the triangle ABD, fig. 48. then the angle DAB is mea-
fured by half the arch BKD (by cor. 1 of art. 62.)
but the half of BD, viz. BE, is thq.fine of half the arch
BKD, viz. the fine of BK (by cor. 2. of the lafl) which
is the meafure of the angle BAD ; therefore the half of
BD is the fine of the angle BAD; the fame way it may
be proved, that the Ijalf of AD is the fine of the angle
ABD, and the half of AB is the fine of the angle
ADB.
66. The fine, tangent, 6^. of -any arch is called alfo
the fine, tangent, &c. of this angle whofe meafure the
arc is : thus becaufe the arc'GD (fig. 49.) is the meafure
of the angle GCD; and fence GH is the fine, DE the
tangent, HD the verfed fine, CE the fecant, alfo GK
the co-fine, BF the co tangent, and CF the co-fecant,
&c. of the arch GD ; then GH is called the fine, DE
the tangent, &c. of the angle GCD, whofe meafure is
the arch GD.
6’j.J/t’ivo equal and parallel lines, AB andCD (fig.JO.)
be joined by two others, AC and BY) ; then thefe Jhall al¬
fo be equal and parallel. Todemonftrate this, join thetwo
oppofite angles A and D with the line AD; then it is
plain this line AD divides the quadrilateral, ACDB, into
two triangles, viz. ABD, ACD, in which AB a leg of
the one, is eqijal to DC a leg of the other, by fuppofiti-
on, and AD is common to both triangles; and fince AB
is parallel to CD, the angle BAD will be equal to the
.angle ADC, (by art. 36.) therefore in the two trian¬
gles BA and AD ; and the angle BAD is equal
to CD and DA; and the angle ADC, that is, two
legs and the included angle in the one, is equal to two
'legs and the included angle in the other ; therefore (by
■art. 61 ) BD is equal to AC, and fence the angle DAC
Vol. H. No. .54. 2
E T H Y. #8$s
iis equal to the angle ADB, therefore the’lines BD AC
are parallel (by w. art. jq.)
Cor. .1. Hence it is plain, that the quadrilateral ABDC
is a parallelogram, fence the oppofite fides are parallel.
Cor. 2. In any parallelogram the line joining the op¬
pofite angles (called the diagonal) as AD, divides the fi¬
gure into two equal parts, fence it has been proved that
the triangles ABD ACD are equal to one another.
■Cor. 3. It follows alfo, that a triangle ACD on the
fame bafe CD, and -between the fame parallels with
a parallelogram .ABDC, is the half of that parallelo¬
gram.
Cor..q. Hence it is plain, that the oppofite fedes of a
parallelogram are equal ; for it has been proved, that
ABDC being a parallelogram, AB will be equal to CD,
and AC equal to BD.
68. All parallelograms on the fame or equal bafes,
and between the feme parallels, are-equal to one another ;
•that is, if BD and GH (tig. 51.) be equal, and the lines
BH and AF be parallel, then the parallelograms ABDG,
BDFE, and EFHG, are equal tu one another. For AC
is equal to EF, each being equal to BD, (by cor. 4. of
67.) To both add CE, then AE will be equal CF. So
in the two triangles ABE CDF, AB a leg of the one,
is equal to CD a leg in the other; and AE is equal to
CF, and theangleBAE is equal to the angle DCF (by art.
• 36.) ; therefore the two triangles ABE CDF are e-
-qual (by art. 6i.)'; and taking the triangle CKE from
both, the figure ABKC will be equal to the figure KDFE ;
to both which add the little triangleKBD, then the paral¬
lelogram ABQC will be equal to the parallelogram BDFE.
The fame way it may be proved, that the parallelogram
EFHG is equal to the parallelogram EFDB ; fo the
three parallelograms ABDC, BDFE, and EFHG will
be equal to one another.
Cor. Hence it is plain, that triangles on the fame bafe,
and between the fame parallels, are equal; fince they are
the half of the parallelograms on the fame bafe and be¬
tween the fame parallels. (by cor. 3. of h(Y art.)
69. In any right-angled triangle, ABC, (fig. 52.)
the fquare of the hypothenufe BC, viz. BCMH, is -qua!
to the fum of the fquares made on the two fides AB and
AC, viz. /<? ABDE and ACGF. To demonftrate this,
through the point A draw AKL perpendicular to the
hypothenufe BC, join AH, AM, DC, and EG; then
it is plain that DB is equal to BA (by art. 53.), alfo BH
is equal to BC (by the fame); fo in the two triangles
DBC ABH, the two legs DB and BC in the one are
equal to the two legs AB and BH in the other; and the
included angles DBC and ABH are alfo equal; (for DBA
is equal to CBH, being both right; to each add ABC,
then it is plain that DBC is equal to ABH) therefore the
triangles DEC ABHareequal(by art. 61.) butthe triangle
DBC is h&lf of the fquare ABDE (by cw. 3. of 67th)
ancl the triangle ABH is half the parallelogram BKLH
(by the fame), therefore half the fquare ABDE is equal
to half the parallelogram BKL. Confequently the fquare
ABDE is equal to the parallelogram BKLH, The fame
way it may be proved, that the fquare ACGF is equal
to the parallelogram KCML. So the fum of the fquares
ABDE and ACGF is equal the fum of the parallelo¬
grams BKLH and KCML, but the fum of thefe paral-
lellograms is equal to the fquare_ BCMH, therefore the
5.L ittis