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Fig. 22.

Fig. 23.

Fig. 24.

fore falls upon the point F, and the two triangles coin.
cide in all their parts; they are not different, therefore,
but equal; and we say, universally, When two triangles
have the three sides of the one equal to the three sides of
the other respectively, the angles will also be equal, re-
spectively, and the two triangles will be cqual in all their
parts. This is the first case of equal triangles.
40. Problem. Two sides of a triangle being given,
and the angle contained by these sides, to construct the
triangle. r
Let A and B (fig. 23) be the two given sides, and C the
given angle. Draw DE equal to the given line A ; at D
make the angle EDG, equal to the give angle C, (32)
and produce the side DG to F, making DF equal to the
given line B; then join FE and you have the triangle
required.
41. It is evident that no different triangle can be con-
structed with these two sides containing an angle equal
to C. The only different construction with these things
given, would be to make the triangle on the other side
of the base DE; but it is plain that the triangle thus
formed would only be the triangle DEF turned over ; the
two triangles would therefore be equal in all their parts.
And we say universally,– When two triangles have two
sides of the one equal to two sides of the other, each to
each, and the angle contained by these two sides of the
one, equal to the angle contained by the two sides of the
other, the two triangles are equal in all their parts. This
is the second case of equal triangles.
42. Problem. One side of a triangle, and the two an-
gles adjacent to that side, being given, to construct the
triangle.
Let the line A (fig. 24) be the given side, and the an-
gles B and C, be the angles adjacent to that side. Draw
the line DE equal to A, for the base of the triangle ; then
at the point D, make an angle equal to the given angle B ;
and at the point E, make another angle on the same side
of DE, equal to the other given angle C ; produce these
two sides till they meet in F, and the triangle is con-
structed.
43. It is manifest that no different triangle can be
formed with these three things given. We may, as in
the other cases, construct the triangle on the lower side
of the base DE; but the triangle thus formed might be

+...or " turned over, and flaced upon the triangle DEF, and because the bases are equal, and the angles at the bases the same, the side DF' would fall upon DF, and the side EF would fall upon EF : the point F" would therefore fall directly upon the point F : and the two triangles would entirely coincide in all their parts. We say, therefore—If two triangles have a side and the two adjacent angles of the one, equal to a side and the two adjacent angles of the other, each to each, the two triangles will be equal in all their parts; that is, the third angle of the one will be equal to the third angle of the other, and the two remaining sides of the one will be equal respectively to the two remaining sides of the other. This is the third case of equal triangles. 44. Problem. The hypothen use and one other side of a right-angled triangle being given, to construct the triangle. Y Let the line A (fig. 25) equal the hypothenuse, and the line B, the other given side of the triangle. Draw the line CD equal to the given side B ; and at D draw the indefinite line DE, perpendicular to CD ; then from C as a centre, with a radius equal to the given hypothen use, draw an arc cutting the perpendicular in E ; draw CE, and you have the right-angled triangle required. [Let the learner show that no different triangle could be constructed with these things given.] We say then—If two right-angled triangles have “he hypothen use and a side of the one, equal respectively to the hypothen use and a side of the other, the two triangles will be equal in all their parts. This is the fourth case of equal triangles. 45. We have seen that the sum of the three angles of every triangle, is equal to two right-angles; yet there is an infinite variety in the magnitude of the individual ansties; and this evidently depends upon the relative maginitudes of the sides. We have learned that when the sides are the same, the angles must be the same (39). Let us examine this subject a little farther. We will

Fig. 25.

take the isosceles triangle ABC (fig. 26), of which AB Fig. 26.

and AC are the two equal sides. From the vertex A, to

the middle of the base BC, draw the straight line AD.

This will divide the triangle ABC, into two trian

gles ADB, ADC. If we examine them, we shall see

that AB of the one is equal to AC of the other, that DB 2

Fig. 26.

Fig. 27.

of the one is equal to DC of the other, and that AD is a
side common to both ; they are therefore equal by the
first case of equal triangles ; and if ABD were turned
over upon ADC, we should find that the two would co-
incide in all their parts. The angle B is therefore equal
to the angle C ; the two angles whose vertices are at
A, equal to each other ; and the two angles whose
vertices are at D, equal to each other ; the two last
are right angles of course (11). Whence we derive
the following important truths—l. In an isosceles, tri-
angle, the angles opposite to the equal sides, are equal.
2. A straight line drawn from the summit of the isosce-
les triangle to the middle of the base, is perpendicular to
the base, and bisects the angle whose vertea is at the sum-
zmit.
46. If the triangle had been equilateral (fig. 27), that
is, ha". had its three sides all equal, either side might
have been taken for the base. This would have proved
the angle A equal to the angle B, and the angle C equal
to the angle B ; hence—An equilateral triangle is also
equiangular ; that is—has all its angles equal.
As the sum of the three angles of every triangle,
is equal to two right-angles, or 180 degrees; in an
equilateral triangle, cach angle is an angle of 60 degrees.
If you have the angle whose vertex is at the summit of
an isosceles triangle, how can you find the other angles 7
If you have an angle at the base of an isosceles trian-
gle, how will you find the other angles
47. Take next the scalene triangle ABC (fig. 28). Let

e AB be the greater side ; take upon AB, the part AD

equal to AC, and draw DC. The angle ADC may be considered as an exterior angle to the triangle DBC, and therefore greater than the angle B; the triangle ADC is isosceles, and consequently has the angle ACD equal to the angle ADC ; but the angle ACB is greater than ACD ; for a still stronger reason is the angle ACB, greater than ABC. But the side AB is greater than AD, or than its equal AC. We see then that the greater side is opposite to the greater angle. If we had compared either of these angles with the angle A, and also their opposite sides, the result would have been similar ; whence we have this general rule—In any triangle, the greater side is opposite to the greater angle. The side opposite to an angle is frequently said to subtend the angle.

48. If two sides of a triangle are unequal, the opposite angles will be unequal (47); therefore——If in any triangle two of the angles are equal, the sides opposite those angles will be equal, and the triangle will be isosceles. 49. In a right-angled triangle, as the two oblique angles must be each less than a right-angle, –the hypothenuse (the side opposite the right angle) must be greater than either of the other sides. The two sides which contain the right-angle are frequently called the legs of the triangle. 50. Either side of a triangle, being a straight line and therefore the shortest which can be drawn between the vertices at its extremities, must be less than the sum of the other two sides. It follows from this, that either side of a triangle must be greater than the difference between the other two sides; if this were not the case, the greatest would exceed or at least equal the sum of the two others. ... 51. Suppose AB (fig. 29) perpendicular to DC, and Fig. * AE and AF two lines drawn from the same point A, and falling obliquely upon DC at equal distances from the point B; the two triangles ABE and ABF, will have AB a common side, EB of the one equal to BF of the other, and the angle ABE equal to the angle ABF (11); the triangles will be equal by the second case, and will have the hypothen uses AE, AF equal, and each greater than the perpendicular AB (49). Draw AG at a greater distance from the perpendicular than AE is ; you will have the triangle AEG obtuse-angled at E, because this angle is the supplement of the acute angle AEB (28); and in the triangle AEG, the side AG being opposite to the obtuse angle is greater than AE opposite an acute angle ; and we have these three propositions. If from a point without a straight line, a perpendicular be drawn to that line, and also several oblique lines, – (1.) The perpendicular is less than either of the oblique lines. (2.) Oblique lines equally distant from the perpendicular are equal ; and (3.) Of two oblique lines, that is the greater which is at the greater distance from the perpendicular. 52. It follows from the second of these propositions, that—If a perpendicular be drawn through the middle of a line, A B, every point in the perpendicular will be equally distant from the two extremities of this line; for

Fig. 29,

Fig. 30.

Fig. 31.

straight lines drawn from any point in the perpendicular
to the two extremities, will be oblique lines drawn at
equal distances from the perpendicular, and therefore
equal.
53. As two points are sufficient to fix the position of
a straight line ; If one straight line pass through the
middle of another straight line [as AB through the mid-
dle of EFJ; and if any other point in the first of these
straight lines, be at equal distances from the two extremi-
ties of the second, these lines will be perpendicular to cach
other. And for the same reason——If each of two points
in a straight line, be equally distant from the two ex-
tremities of another straight line, the first will bisect the
second, and be perpendicular to it. To bisect is to divide
into two equal parts.
54. Problem. From any point in a given straight line,
to draw a perpendicular to that line.
Let A be a given point in the line BC (fig. 30), from
which it is proposed to draw a perpendicular. Take two
points B and C in the given line, equally distant from
the given point A ; from each of these points B and C,
with any radius greater than AB, draw arcs cutting each
other in some point D, above or below the given line ;
and from A, draw a line through D, which will be the per-
pendicular required (53).
It is evident that there can be only one line per-
pendicular to BC, and passing through the point A ; for
if there could be two, as they make the same angle with
the line BC, they will both have the same direction ; and
having the same direction, and passing through the same
point, they must coincide entirely, and be one and the
same straight line. There cannot, therefore, be more
than one perpendicular drawn through the same point in
the same straight line.
55. Problem. From a given point without a straight
line, to draw a perpendicular to that line.
Let A (fig. 31) be the given point, and BC the given
straight line. From A as a centre, with any radius
greater than the distance of the point A, from the given
line, draw an arc cutting the given line in the points
D and E ; then from each of these two points D and E,
with a radius greater than half of DE, draw arcs cut-
ting each other in F; and from the point A, in the direc-
tion AF, draw the line AG; this line will be the per-

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