54. Corollary. From the preceding three theorems, it follows that an angle of a triangle is acute, right or obtuse, according as the square of the side opposite to it is less than, equal to, or greater than, the sum of the squares of the other two sides.

PROPOSITION XVII.--THEOREM.

55. If through a fixed point within a circle any chord is drawn, the product of its two segments has the same value, in whatever direction the chord is drawn.

Let be any fixed point within the circle 0, AB and A'B' any two chords drawn through P; then,

For, join AB' and A'B. The triangles APB', A'PB, are similar, having the angles at P equal, and also the angles A and A' equal (II. 58); therefore,

A A'

B

B

whence (5),

PA: PA' =

: PB' : PB,

PAX PB PA' × PB'.

=

=

56. Corollary. If AB is the least chord, drawn. through P (II. 20), then, since it is perpendicular to OP, we have PA PB (II. 15), and hence PAPA' X PB'; that is, either segment of the least chord drawn through a fixed point is a mean

proportional between the segments of any other chord drawn through that point.

57. Scholium. If a chord constantly passing through a fixed point P, be conceived to revolve upon this point as upon a pivot, one segment of the chord increases while the other decreases, but their product being constant (being always equal to the square of half the least chord), the two segments are said to vary reciprocally, or to be reciprocally proportional (2).

PROPOSITION XVIII.-THEOREM.

58. If through a fixed point without a circle a secant is drawn, the product of the whole secant and its external segment has the same value, in whatever direction the secant is drawn.

Let P be any fixed point without the circle 0, PAB and PA'B' any two secants drawn through P; then,

PAX PB: = PA' × PB'.

For, join AB' and A'B. The triangles APB', A'PB, are similar, having the angle at P common, and also the angles B and B' equal (II. 58); there

fore,

PA: PA' PB': PB,

=

T

P

B

B'

whence (5),

PAX PB

=

PA' × PB'.

59. Corollary. If the line PAB, constantly passing through the fixed point P, be conceived to revolve upoǹ P, as upon a pivot, and to approach the tangent PT, the two points of intersection, A and B, will approach each other; and when the line has come into coincidence with the tangent, the two points of intersection will coincide in the point of tangency T. The whole secant and its external segment will then both become equal to the tangent PT; therefore, regarding the tangent as a secant whose two points of intersection are coincident (II. 28), we shall have

that is, if through a fixed point without a circle a tangent to the circle is drawn, and also any secant, the tangent is a mean proportional between the whole secant and its external segment.

60. Scholium I. When a secant, constantly passing through a fixed point, changes its direction, the whole secant and its external' segment vary reciprocally, or they are reciprocally proportional, since their product is constant (2).

61. Scholium II. The analogy between the two preceding propositions is especially to be remarked. They may, indeed, be reduced to a single proposition in the following form: If through any fixed

point in the plane of a circle a straight line is drawn intersecting the circumference, the product of the distances of the fixed point from the two points of intersection is constant.

PROPOSITION XIX.-THEOREM.

62. In any triangle, if a medial line is drawn from the vertex to the

base:

1st. The sum of the squares of the two sides is equal to twice the square of half the base increased by twice the square of the medial line; 2d. The difference of the squares of the two sides is equal to twice the product of the base by the projection of the medial line on the base. In the triangle ABC, let D be the middle point of the base BC, AD the medial line from A to the base, P the projection of A upon the base, DP the projection of AD upon the base; then,

1st. AB+ AC2 = 2BD' + 2AD';

Α

2d. AB-AC2BC X DP.

C

For, if AB > AC, the angle ADB will be obtuse and ADO will be acute, and in the triangles ABD, ADC, we shall have, by (53) and (52).

AB2 = BD2 + AD2 + 2BD × DP,

AC2= DC2 + AD2 — 2DC × DP.

Adding these equations, and observing that BD = DC, we have

1st. AB2 + AC2 = 2BD2 + 2AD2.

Subtracting the second equation from the first, we have

63. Corollary I. In any quadrilateral, the sum of the squares of the four sides is equal to the sum of the squares

of the diagonals plus four times the square of the line joining the middle points of the diagonals.

For, let E and F be the middle points of the diagonals of the quadrilateral ABCD; join EF, EB, ED. Then, by the preceding theorem, we have in the triangle ABC,

AB2 + BC2 = 2AE2 + 2BE3,

and in the triangle ADC,

CD2+ DA2 = 2AE2 + 2DE2,

whence, by addition,

B

E

AB2+ BC2 + CD2 + DA2 = 4AE2 + 2 (BE2 + DE3).

Now, in the triangle BED, we have

therefore,

BE+DE=2BF2 + 2EF";

2

AB2 + BC2 + CD + DA2 = 4AE2 + 4BF2 + 4EF'.

AB2+ BC2 + CD2 + DA2 = AC2 + BD2+4EF2.

:

64. Corollary II. In a parallelogram, the sum of the squares of the four sides is equal to the sum of the squares of the diagonals. For if the quadrilateral in the preceding corollary is a parallelogram, the diagonals bisect each other, and the distance EF is zero.

PROPOSITION XX.-THEOREM.

65. In any triangle, the product of two sides is equal to the product of the diameter of the circumscribed circle by the perpendicular let fall upon the third side from the vertex of the opposite angle.

Let AB, AC, be two sides of a triangle ABC, AD the perpendicular upon BC, AE the diameter of the circumscribed circle; then,

AB X AC= AE × AD.

For, joining CE, the angle ACE is a right

angle (II. 59), and the angles E and B are equal (II. 58); therefore, the right triangles AEC, ABD, are similar, and give

AB: AE AD: AC,

=

whence, AB X AC= AEX AD.

PROPOSITION XXI.-THEOREM.

66. In any triangle, the product of two sides is equal to the product of the segments of the third side formed by the bisector of the opposite angle plus the square of the bisector.

Let AD bisect the angle A of the triangle ABC; then,

ABX AC= DB × DC+ DA'.

DI

For, circumscribe a circle about ABC, produce AD to meet the circumference in E, and join CE. The triangles ABD, AEC, are similar, and give

AB: AE-DA: AC,

D

whence ABX AC= AE X DA = (DE + DA) × DA =DEX DA + DA'.

Now, by (55), we have DEX DA = DB × DC, and hence

67. Corollary. If the exterior angle BAF is bisected by AD', the same theorem holds, except that plus is to be changed to minus.