=

=

b'B' B'c' c'C'.

=

... A'a' = a'b', and for a like reason, a'b'

.. AB BC:: 3 : 2, and A'B': B'C' :: 3 : 2.

.. AB: BC:: A'B': B'C'.

79

The same reasoning will be applicable, it is obvious, whatever may be the number of parts into which AB, BC, may be divided, or whenever AB, BC are commensurable.

a

αν

Next, let the segments a, b, be incommensurable. Produce b so that be shall be commensurable witn a, and through the extremity of a draw a parallel increasing the corresponding segment b' by x'; then will b'+' be commensurable with a'; and, by what has already been demonstrated, we shall have the proportion

Fig. 232.

but x can be made less than any assignable quantity, since it is less than the measure of a, which may be diminished at pleasure;

whence it follows that

x'

a

, are variable quantities capable of

indefinite diminution, and, consequently (63), that

Cor. 1. When three lines, two of which are parallel, are (121) cut by two others so as to make the segments of the secant lines proportional, the third line is parallel to the other two. For, if the line CC' be not parallel to AA', let CC" be so; then there will result (120)

but, by hyp., AB : BC :: A'B': B'C', or B'C'

C"

Fig. 233.

... B'C' = B'C', which is absurd; therefore CC' is not otherwise

than parallel to AA'.

Cor. 2. The sides of mutually equiangular triangles are (122) proportional.

For, let any two of the equal angles be made vertical, then will

the bases m, n, be parallel, and, if through the common vertex a line be drawn parallel to m, n, dividing their intercepted perpendicular into the parts a", b', we shall find (120)

Fig. 23

ab:: a": b' :: a' : b'. Q. E. D. Remark. The homologous sides are opposite equal angles.

Cor. 3. If two triangles have an angle of the one equal to (123) an angle of the other and the sides about the equal angles proportional, the triangles will be mutually equiangular and consequently similar. Draw a line through the extremity of b [fig. 23], see (122), and imitate the reasoning under (121).

Def. Figures which have their angles respectively equal when taken in the same order, and their sides proportional, also taken in the same order, are called similar.

Cor. 4. The altitudes of similar triangles are to each oth- (124) er as their bases. For we have (122)

a" : b" :: a b :: m : n.

Cor. 5. If lines be drawn from the vertex of a triangle to (125) points in the base, all lines parallel to the base will

be divided into parts proportional to the segments of the base.

ABC P

For the segments ab, bc, cd, have to AB, BC,
CD, the constant ratio of the perpendiculars Op, Fig. 235.

OP.

Cor. 6. Triangles having their sides severally parallel are (12) similar.

نا

Fig. 236.

C

Cor. 7. Triangles which have their sides respectively per- (127) pendicular, are similar. Thus, a, b, c, are to each other as a', b', c', where it is obvious that the homologous sides are those that are perpendicular to each other.

a

A right angled triangle, by a perpendicular let fall from (128) the right angle, is divided into two partial triangles, similar to itself, and consequently to each other.

HYPOTHENUSE.

a P

m

h

n

81

b

Let a, b, denote the sides about the right angle, h the side opposite, or the hypothenuse, and m, n, the segments of h made by the perpendicular, p. The angle (a, h) is common to the triangles (a, b, Fig. 24. h), (a, m, p), which also have a right angle in each, and are therefore equiangular, (≤ (b, h) being = ≤ (p, a),) and consequently similar. So it may be shown that A (b, n, p) is similar to ▲ (h, b, a), and the proposition is demonstrated.

Cor. 1. The perpendicular is a mean proportional be- (129) tween the segments of the hypothenuse.

Cor. 2. The sides about the right angle are mean pro- (130) portionals between the hypothenuse and the adjacent segments. For we have (122)

h b
b n

=- and = a m

Cor. 3. The segments of the hypothenuse are to each (131) other as the squares of the sides opposite them; for, from the above we have

Cor. 4. The square of the hypothenuse is equal to the (132) sum of the squares of the other two sides. For we have

Scholium. This proposition, the celebrated 47th of Euclid, is obviously a direct consequence of (122), and with that theorem, which embraces the characteristic property of the triangle, constitutes the working rules of all geometry.

Cor. 5. Conversely, if the square of one side of a tri- (133) angle is equal to the sum of the squares of the other two, the triangle is right angled.

For suppose the triangle (h, a, b) to be such that h2 = a2 + b2; then if on a we construct the triangle (h', a, b') making bb and the (a, b), there will result h2 = a2 + b'2 = a2 + b2 = h2, ... h' = h, and (a, b) = (a, b) = ↳ .

a

Fig. 242.

16

Cor. 6. The diagonal of a square is incommensurable (134) former being to the latter as 2 to 1.

with its side, the For we have

..

x2 = a2 + a2 = 2a2

x = a√2, or x : a = √2 : 1.

Fig. 243.

The student might very naturally suppose that a common measure to any two lines could be found by taking the measuring unit sufficiently small; but, from this corollary, it is shown that the contrary may be true, and we are therefore taught the necessity of extending our demonstrations to the case of incommensurability.

PROPOSITION III.

To find the relation between the oblique sides, a, b, of a triangle, the line c drawn from the vertex to the base, and the segments, m, n, of the base made by this line.

Drop the perpendicular p, intercepting the segment x between the foot of p and that of c; we have (132), (8), (9),

a2 = p2 + (m + x)2 = p2 + m2 + 2mx+x2,
b2 = p2 + (n − r)2 = p2 + n2 — 2nx+x2,
c2 = p2 + x2;

.. subtracting the 3d from the 1st, we have

Fig. 25

If in any triangle, a line be drawn from one of the angles, terminating in the opposite side, and the squares of the sides embracing the divided angle be severally diminished by the square of the dividing line, then the sum of the quotients arising from dividing these remainders by the corresponding subjacent segments, will be equal to the divided side.

If we take the particular case in which m and n are equal, there will result

Cor. 1. If a line be drawn from the vertex of a triangle (136) to the middle of the base, the sum of the squares of the oblique sides will be double the sum of the squares of the middle line and the half base.

Cor. 2. In an isosceles triangle, the square of the oblique (137) side diminished by the square of the middle line, is equal to the product of the segments of the base.

Let ABCD be any quadrilateral, M, N, the middle points of its diagonals, AC, BD; join AN, NC;

D

M

Fig. 252.

AB+ BC+CD2 + DA2 = 4BN2+2(AN2+CN2)

==

=4BN2+2(2AM2 + 2MN2)

= (2AM)2+(2BN)2 + 4MN2 =AC2 + BD2+4MN2.

Cor. 3. The sum of the squares of the sides of a (138) quadrilateral, exceeds the sum of the squares of its diagonals by four times the square of the line joining the middle points of the diagonals.

Cor. 4. The sum of the squares of the sides of a parallelo- (139) gram, is equal to the sum of the squares of its diagonals. For (99) the diagonals bisect each other.

PROPOSITION IV.

To find the distance of the foot of the perpendicular of a triangle from the middle of its base.

Let a, b, be the oblique sides of a triangle, and p its perpendicular, dividing the base 2m into two parts, m+x, m