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ADOA={D0.OA sin 0, and adding,

Qd. = JAC.BD sin 0.

(compare 162°)

B

B

...

217o. BD being the altitude to AC in the AABC, we have

from 172°, 2,

a = b + c2 – 26. AD. But AD=AB cos A=c cos A,

a-=62 +62 — 2bc cos A.

When B comes to B' the LA becomes obtuse, and cos A changes sign.

(214°, 2) If we consider the cosine with respect to its magnitude only, we must write + before the term abc cos A, when A becomes obtuse. But, if we leave the sign of the function to be accounted for by the character of the angle, the form given is universal.

D

A

D

с

B

с

a

А

P

b

E

Cor. 1. ABCD is a parallelogram. Consider the AABD,

then BD’=a2 + 62 2ab cos 0.

Next, consider the AABC. Since LABC is the supplement of 0, and

BC=AD=b,

AC=m2 +62 + 2ab cos 0. and writing these as one expression,

a + b2+ 2ab cos 0, gives both the diagonals of any one of whose angles is 0.

DE=a cos 0 (CE being I to AD), CE=a sin 0.
AE=6+ a cos 0 ;

CE and

a sin 0 tan CAE

AE6+ acos O' which gives the direction of the diagonal.

Cor. 2.

218o. Def.The ratio of any area X to another area Y is the measure of X when Y is taken as the unit-area, and is

X accordingly expressed as (Compare 1889.)

Y 1. Let X and Y be two similar rectangles. Then X=ab

and Y=a'b', where a and b are adjacent sides of the OX and a' and b' those of the OY.

X 6
Υ α''

b But because the rectangles are similar,

a

6° X na

Y a'2 i.e., the areas of similar rectangles are proportional to the areas of the squares upon homologous sides.

2. Let X and Y be two similar triangles. Then

X=}ab sin C, Y=}a'sin C,
X ab a2

Ya'w a'2 because the triangles are similar,

(197°) i.l., the areas of similar triangles are proportional to the areas of the squares upon homologous sides.

B

B

R

E

P_R_Q=

=

3. Let X denote the area of the pentagon ABCDE, and Y that of the similar pentagon A'B'C'D'E'. Then

R
P AD2 R AC?
PA'D' R'A'C'29
Q. DC

Q D'C"2
But

(X)

(Y) P+Q+R X

(195°, 3) P

R' Q' P'+Q+R' Y
AD AC DC

(207°) A'D' A'C' D'C'

X DC? ..

Y D'C"2 And the same relation may be proved for any two similar rectilinear figures whatever.

.. the areas of any two similar rectilinear figures are proportional to the areas of squares upon any two homologous lines.

and,

4. Since two circles are always similar, and are the limits of two similar regular polygons,

.. the areas of any two circles are proportional to the areas of squares on any homologous chords of the circles, or on line-segments equal to any two similar arcs.

5.. When a figure varies its magnitude and retains its form, any similar figure may be considered as one stage in its variation.

Hence the above relations, 1, 2, 3, 4, may be stated as follows:

The area of any figure with constant form varies as the square upon any one of its line-segments.

EXERCISES.

1. Two triangles having one angle in each equal have

their areas proportional to the rectangles on the sides

containing the equal angles. 2. Two equal triangles, which have an angle in each equal,

have the sides about this angle reciprocally proportional, i.e.,

a: a'=': 6. 3. The circle described on the hypothenuse of a right-angled

triangle is equal to the sum of the circles described on

the sides as diameters. 4. If semicircles be described outwards upon the sides of a

right-angled triangle and a semicircle be described inwards on the hypothenuse, two crescents are formed

whose sum is the area of the triangle. 5. AB is bisected in C, D is any point in

AB, and the curves are semicircles.

Prove that P+S=Q+R. 6. If a, b denote adjacent sides of a parallelogram and also

of a rectangle, the ratio of the area of the parallelogram to that of the rectangle is the sine of the angle of the parallelogram.

Q

S

А

с

D

B

7. The sides of a concyclic quadrangle are a, b, c, d. Then the cosine of the angle between a and b is

(a* + b2 --d)2(ab+cd). 8. In the quadrangle of 7, if s denotes one-half the perimeter,

area= {(s-a)(s --6)(s - c)s--d)}. 9. In any parallelogram che ratio of the rectangle on the

sum and differences of adjacent sides to the rectangle on the diagonals is the cosine of the angle between the

diagonals. 10. If a, b be the adjacent sides of a parallelogram and the

angle between them, one diagonal is double the other when cos 0=

3a

10 6 u. If one diagonal of a parallelogram is expressed by

na + 1 12. Construct an isosceles triangle in which the altitude is a

mean proportional between the side and the base. 13. Three circles touch two lines and the middle circle

touches each of the others. Prove that the radius of the middle circle is a mean proportional between the

radii of the others. 14. In an equilateral triangle describe three circles which

shall touch one another and each of which shall touch

a side of the triangle. 15. In an equilateral triangle a circle is described to touch

the incircle and two sides of the triangle. Show that its radius is one-third that of the incircle.

M

PART IV.

SECTION I.

GEOMETRIC EXTENSIONS.

M

B

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220°. Let two lines L and M passing through the fixed points A and B meet at P. When P moves in the direction of the arrow, L and M

approach towards parallelism, and the angle APB diminishes. Since the

lines are unlimited (21°, 3) P may recede from A along L until the segment AP becomes greater than any conceivable length, and the angle APB becomes less than any conceivable angle.

And as this process may be supposed to go on endlessly, P is said to "go to infinity” or to “be at infinity," and the LAPB is said to vanish.

But lines which make no angle with one another are parallel, .. Parallel lines meet at infinity, and lines which meet at infinity are parallel.

The symbol for “infinity” is co.

The phrases “to go to infinity," "to be at infinity,” must not be misunderstood. Infinity is not a place but a property. Lines which meet at oo are lines so situated that, having the same direction they cannot meet at any finite point, and therefore cannot meet at all, within our apprehension, since every point that can be conceived of is finite.

178

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