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A c' c c' C c'.
A b/T b b/ . B 5' "
straight lines are cut by any number of parallel lines they
are divided into proportional parts; and if one of the
lines is divided into equal parts, the parts of the other
will be equal.
97. To find a fourth proportional to three given lines
(fig. 46). Draw the line AB equal to the sum of the
first and second lines in the proportion, take AD equal to
c c' e ; therefore, W = that is—When two
the first ; and draw AC equal to the third, and making any
angle with AB; join DC, and through B draw BE paral-
lel to DC; produce AC to E ; because DC is parallel to
BE, we have ; - ; (96); therefore CE is the fourth
98. To find a third proportional to two given lines.
Take a third line equal to the second ; and find the
fourth as above ; this will be the third proportional re-
quired. In this case the second of the given lines is
called a mean proportional between the first and third.
99. Let ABC (fig. 47) be a right-angled triangle; from
the vertex A of the right angle draw the perpendicular
AD to the hypothenuse BC. If we compare the triangle
ABD with the original triangle, we shall see that they
have each a right-angle, and the common angle B; the
other angles must therefore be equal (35), viz. the an-
gle C to the angle BAD ; the triangle ABD is therefore
similar to the triangle ABC ; in the same manner it may
be shown that the other triangle ADC is similar to the
triangle ABC (91); whence we say—If from the vertex
of the right-angle of a right-angled triangle, a straight
line be drawn perpendicular to the hypothenuse, it will di-
vide the triangle into two triangles, each similar to the
100. The similar triangles ABD, ACD, give the pro-
..., BD AD . that is—Th dicular i
portion AD T DC * at 1s—I he perpenalcular as a
mean proportional between the two segments of the hypoth-
enuse. The similar triangle ABC, ABD, give the pro-
BA. T BD
between BC and BD. In the same manner we could
show that CA is a mean proportional between BC and
portion that is, BA is a mean proportional
LC; therefore—Each side of the original triangle is a
mean proportional between the hypothenuse and the seg-
ment of the hypothemuse adjacent to that side.
We shall hereafter show the method of finding " to on
proportional between any two given lines.
e BD BA o * (BA)? .
101. The equation |BA. T BC’ gives BD = TECT * . DC A . CA)2
and the equation GA = BC, gives DC — o ad
ding these two equations we obtain BD + DC = BC =
(BA)* + ( ) ; multiplying both sides by BC, we have.
(BC)* = (BA)2 + (AC)”. From this it is evident that if we refer to a common measure the three sides of a right-angled triangle, the second power of the number expressing the length of the hypothen use, will be equal to the sum of the second powers of the numbers which express the lengths of the other two sides. We can therefore find the hypothen use when we have the other sides. Suppose in this triangle that AB contains the common measure six times, and AC eight times; then (BC)* = 36 -–64 = 100; and taking the second root of
I both sides of this equation, we have BC = (100)* = 10. So also if we know the hypothen use and one of the sides, we can find the other. Suppose we have the hypothenuse equal to 10, and the side AB equal to 6, to find the side AC. The equation (BC)2 = (BA)2 + (AC)”, by subtracting (BA)* from both sides, will give (AC)* = (BC)” – (BA)2 = 100 — 36 = 64; and taking the sec
ond root of each side, we obtain AC = (64)? = 8.
102. In the triangle ABC (fig. 48), bisect the angle B by the line BD; and through the point C, draw CE parallel to DB till it meets AB produced. The angle ABC, exterior to the triangle CBE, is equal to the sum of the two interior opposite angles BCE and BEC (37); but on account of the parallels DB and CE, the angle ABD (half of ABC) is equal to the angle E.; the other half must be equal to the angle BCE, consequently BCE and BEC will be equal ; therefore, the opposite sides BC and BE are equal (45). Now in the triangle AEC, DB is drawn parallel to the base, which
3. # = ; (98); but BC is equal to BE, and may be substituted for it in the proportion; AB AD .
BC T DC * secting an angle of a triangle, divides the opposite side into parts proportional to the adjacent sides.
which gives, that is—A straight line bi
103. Let us now take an obtuse-angled triangle, as
ABC (fig, 49) and draw perpendiculars from the verti-
ces of each of the angles to the opposite sides. The
lines AF and BG, respectively perpendicular to BC and
CA, will meet in the point D ; we wish to ascertain the
point M, in which CE perpendicular to BA, meets BG.
First, it is manifest that, as the two triangles ABG
and EBM, have each a right angle, and the angle EBM
common, they are equiangular and similar ; and give the
BG T BE’
If we now compare the triangles EBC, and ABF, hav-
ing each a right angle and the common angle ABC, they
give the proportion o-- BA. And a comparison of
the two triangles BGC, BFD, give the proportion
BC × BD BC × BA.
BFox BET BFX, BG
ratio of this proportion by the common multiplier
have BG T BE *
the first BA BM .
he ursi, BG = BE;
. BA e
common ratio Bā; the other two ratios are therefore equal,
and give the proportion ; - #: that is, BM is equal
to BD, and the three perpendiculars meet at the same
If the original triangle were acute-angled, as ABD
in the same figure, it is manifest that the three perpen-
Multiplying in order, the two last proportions,
gives if now we divide each
and comparing this proportion with
we perceive that they have the
diculars drawn from the vertices to the opposide sides, would meet in the point C. We say therefore—In any triangle, the perpendiculars drawn from the vertices of the angles to the opposite sides, will meet in the same oint.
p Remark. If the triangle be acute-angled, the point of meeting will be within the triangle ; if the triangle be obtuse-angled, the point of meeting will be without ; and in the right-angled triangle, the point of coincidence is the vertex of the right-angle. '
Of the Straight Line and Circle.
104. The circle is the plane surface embraced by the circular curve already described (7), called the circumference of the circle. A straight line which touches the circumference only in one point, is called a tangent, as AB (fig. 50). When it is said that the straight line touches the circumference, it is meant that its direction is such that if it were produced ever so far each way, it would have only one point in common with the circumference. This point is called the point of contact. -
105. If the straight line cuts the circumference in two points, it is called a secant. A straight line which has its two extremities in the circumference, as DE (fig. 51), is called a chord; it subtends both the arc and the angle at the centre measured by this arc. The portion of surface contained between the two radii DO, EO, and the arc DGE, is called a sector. The portion contained between the chord and the arc is called a segment of the circle.
106. If the two radii DO, EO, the chord DE and the arc DGE, be made to revolve together about the centre O, of the circle, the points D and E will be continually in the circumference ; and the arc DEG will always coincide with the part of the circumference then embraced by the two radii (7); if this were not the case, some parts of the circumference must be nearer to the centre than other parts; moreover, the straight line DE would be constantly the chord of this arc. Now as the angle DOE would not be changed by this revolution of the sector DOE, we say that this angle would have the same arc in every part of the circumference, and this arc would also have the same chord. We have, therefore,
these propositions:—(1). In the same circle or in equal
circles, equal angles at the centre are measured by equal
arcs ; and equal arcs subtend equal angles. (2). In the
same circle, or in equal circles, equal arcs are subtendel'
by equal chords; and when, in the same circle, the chord,
are equal, the arcs are equal.
As the triangle would not be changed by this revolu-
tion, the distance of the vertex O from the opposite side
DE, would not vary; that is—In the same circle, equal
chords are at the same distance from the centre.
107. If the arc were increased, the angle at the centre
would be increased (26); but the two sides of the trian-
gle OD and OE remaining the same, if their angle is in-
creased, the third side DE must be increased (32); there-
fore—— The greater arc is subtended by the greater chord.
By increasing the side DE of the triangle, we increase
the angle O, and the arc which measures this angle ;
therefore—The greater chord subtends the greater arc.
108. We said that the chord increases as the arc is in-
creased ; this is true to a certain extent; that is, while the
arc is less than 1809, or a semi-circumference. At this
point the chord is confounded with the two radii, and be-
comes a diameter (7); it subtends no angle, and is no
longer called a chord. But as the chord continues to
increase till it is confounded with the diameter, we see
that—Of two unequal chords, the greater is at the less
distance from the centre.
If the arc be increased beyond 1800, the two radii
will make an angle with each other on the other side,
and the chord which subtends this arc, will make the
third side of the triangle, and is therefore less than the
sum of the other two (50), and consequently less than
the diameter. Therefore—In the same circle a chord is
always less than the diameter.
It will be perceived that every chord corresponds to two
arcs, which taken together constitute the circumference.
#39. In fig, 52, suppose the radius CD to be drawn
perpendicular to the chord AB ; if we draw the other
two radii CA and CB, we shall have the isosceles trian-
gle ACB ; CE perpendicular to the base AB, bisects the
base and the angle at C (45); as the two angles at C
are equal, the arcs AD and DB are also equal (107);
whence we say—Radius perpendicular to a chord, bisects
the chord and its arc.