Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

For Euclid I. 30, see Appendix.

PROP. 30. THEOR.

Straight lines which are parallel to the same straight line are parallel to one another. Given AB || CD, and EF | A CD.

с

D To prove AB || EF. If AB is not || EF, they will E

F meet; then there will be two intersecting lines both || CD, which is impossible (Ax. 11).

:: AB is || EF. Q.E.D.

[ocr errors]

PROP. 31. PROB.

Through a given point to draw a straight line parallel to a given straight line. Given st. line AB and pt. C.

E

F To draw through C a st. line || AB.

Ą Take any point D in AB,

and join CD. At C in CD make the L ECD = < CDB (I. 23). Produce EC. :: _ECD

= _ CDB, and these are alternate angles, :: EF is || AB (I. 27), and has been drawn through

the given point C. Q.E.F. 73. Prove Prop. 30 by drawing a line to intersect the given lines.

74. If a straight line be parallel to two straight lines which meet a point, these two shall be in the same straight line.

PROP. 32. THEOR. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of every triangle are together equal to two right angles. Given A ABC having

A
BC produced.
To prove ACD = L A

E

N

+ LB.

2 L.s.

с

[ocr errors]

Also, LA + LB +

B
LACB
Through C draw CE || AB (I. 31).
:: AC meets the parallels AB, CE, :: LA = LACE

(I. 29).
:: BD meets the parallels AB, CE, .. ECD
(I. 29).
:: LACD

= LA +_B.
To each of these equals add – ACB.

:: LACD + L ACB LA + B + - ACB. But ACD + L ACB 2 Ls. (I. 13).

:: LA + 2B + L ACB = 2 L's. Q.E.D. 75. A straight line parallel to the base of an isosceles triangle and intersecting the sides, or the sides produced, forms with them an isosceles triangle.

76. Through a given point to draw a st. line to make with a given st. line an angle equal to a given angle. How many such lines are there?

77. If a straight line be drawn through the vertex of an isosceles triangle parallel to the base, it shall bisect the exterior angle at the vertex.

78. In a right-angled triangle find a point in the hypotenuse AB so that its distance from AC equals its distance from B.

79. In an isosceles triangle find two points in the equal sides equidistant from the extremities of the base and from one another.

[ocr errors]

Cor. 1.—The sum of the interior angles of any rectilineal figure is equal to twice as many right angles as the figure has sides, minus four right angles. Take any rectilineal figure, as

B ABCDEF, and take G, any point within it. Join G with A, B, etc. (1.) There are as many triangles as

G

A the figure has sides.

(2.) The angles in each triangle amount to two right angles (I. 32). (3.) .. the angles in all the triangles

E together amount to twice as many right angles as the figure bas sides (2n (s., n=number of sides).

(4.) But the angles of all the triangles include the angles of the figure at A, B, C, etc., and the angles at G which amount to four right angles.

(5.) .. the angles of the figure amount to twice as many right angles as the figure has sides, minus four right angles (2n L.S. — 4 LI.s.)

.

с

G

Cor. 2.- The sum of the exterior angles of any convex rectilineal figure, formed by producing the sides all in one way, is four right angles.

Take ABCDEF, any convex rectilineal figure having the sides produced all in one way. To prove sum of exterior angles

A 2 four right angles.

(1.) Each exterior angle with its adjacent interior angle, as 41+-2, equals two right angles. (2.) :. all the exterior with all the

F interior amount to twice as many right angles as there are sides in the figure (2n L'.s.) (3.) But all the interior angles together with the angles at amount to twice as many right angles as the figure has

(Cor. 1) (2n _'.s.)
... all the exterior angles with all the interior=all the
ür ang with the angles at G.

E

4

(5.) .. all the exterior angles together= angles at G=four right angles (4 _'.s.)

Note.—When the rectilineal figure is not convex, but has a reflex angle or reflex angles, the result is modified as follows; the proof of which we leave to the student:

21+2+3+4+ 15-16=

4 L.S.

80. The angles of a quadrilateral amount to four right angles.

81. Find the magnitude of the angle in a regular pentagon, hexagon, heptagon, octagon, etc. (Cor. 1).

82. Find the magnitude of the exterior angle of any regular polygon, as a pentagon, nonagon, decagon, or polygon of n sides.

83. The exterior angle of a regular polygon is is of a rt. angle, how

many sides has it? 84. The interior angle of a regular polygon is lrt. angle, how many sides has it?

85. Show that three regular hexagons can be placed so as to have a common angular point and to fill up the space about it.

86. What other regular figures can be formed round common angular points in the same way?

87. Make a pattern composed of octagons and squares.

88. What other regular figures may be combined to fill up the space round the common angular points? Make patterns of the combinations.

89. How many diagonals can be drawn in a pentagon ; in a hexagon ; in a polygon of n sides ?

90. If all the sides of a polygon be produced in both ways, the angles formed by the meeting of every two alternate sides will together amount to twice as many right angles as the figure has sides, minus eight right angles (2n L'.s. — 8 LI.s.)

91. The middle point of the hypotenuse of a right-angled triangle is equally distant from each of the three angles.

92. And conversely, If the middle point of one side of a triangle be equally distant from each of the three angles, the triangle shall be right-angled.

93. Given the perimeter of a triangle and the angles at the base, to construct the triangle.

PROP. 26. THEOR. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, either (1st) the side adjacent to the equal angles in each, or 2nd) a side opposite to one of the equal angles in each, then the two triangles shall be equal in every respect.

B

1. Given triangles ABC and DEF, having

_ B = LE, LC = _F, and BC EF. To prove BA ED, AC DF, and LA = _ D. Place A ABC on A DEF, so that B is on E, and BO

along EF, then C falls on F, :: BC = EF; and BA must lie along ED, : _B LE; and CA must lie along FD, : 20

LF.
:: pt. A must fall on D, and A ABC must coincide
with A DEF.

ED, AC DF, and LA = LD.
А

D

.: BA

A

[blocks in formation]

F

2. Given triangles ABC and DEF, having

LB = LE, LC = _F, and AB = DE.
To prove BC = EF, AC = DF, and _A = _D.

[ocr errors]
« ΠροηγούμενηΣυνέχεια »