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88. If equals be added to equals, the sums are equal.

89. If equals be taken from equals, the remainders are equal.

90. If to equals unequals are added, the sums are unequal, and the greater sum comes from adding the greater magnitude.

91. If equals are taken from unequals, the remainders are unequal, and the greater remainder is obtained from the greater magnitude.

92. Things that are double of the same thing, or of equal things, are equal to one another.

93. Things which are halves of the same thing, or of equal things, are equal to one another.

III. Some Geometrical Assumptions about Euclid's Space.

94. A solid or a figure may be imagined to move about in space without any other change. Magnitudes which will coincide with one another after any motion in space, are congruent; and congruent magnitudes can, after proper turning, be made to coincide, point for point, by superposition.

95. Two lines cannot meet twice; that is, if two lines have two points in common, the two sects between those points coincide.

96. If two lines have a common sect, they coincide throughout. Therefore through two points, only one distinct line can pass.

97. If two points of a line are in a plane, the line lies wholly in the plane.

98. All straight angles are equal to one another.

99. Two lines which intersect one another cannot both be parallel to the same line.

IV. The Assumed Constructions.

100. Let it be granted that a line may be drawn from any one point to any other point.

101. Let it be granted that a sect or a terminated line may be produced indefinitely in a line.

102. Let it be granted that a circle may be described around any point as a center, with a radius equal to a given sect.

103. REMARK. – Here we are allowed the use of a straight edge, not marked with divisions, and the use of a pair of compasses; the edge being used for drawing and producing lines, the compasses for describing circles and for the transferrence of sects.

But it is more important to note the implied restriction, namely, that no construction is allowable in elementary geometry which cannot be effected by combinations of these primary constructions.

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CHAPTER III.

PRIMARY RELATIONS OF LINES, ANGLES, AND TRIANGLES.

I. Angles about a Point.

THEOREM I.

104. All right angles are equal.

Such a proposition is proved in geometry by showing it to be true of any two right angles we choose to take.

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The hypothesis will be, that we have any two right angles ; as, for instance, right X AOB and right % FHG.

By the definition of a right angle (66), the hypothesis will mean, * AOB is half a straight angle at 0, and 4 FHG is half a straight angle at H.

The conclusion is, that 4 AOB and 4 FHG are equal.

The proof consists in stating the equality of the two straight angles of which the angles AOB and FHG are halves, referring to the assumption (in 98) that all straight angles are equal, and then stating that therefore the right X AOB equals the right % FHG, because, by 93, things which are halves of equal things are equal.

Now, we may restate and condense this as follows, using the abbreviations rt, for “right," and st. for "straight," and the symbol .. for the word "therefore":

HYPOTHESIS. * AOB is rt. * ; also X FHG is rt. 4.
CONCLUSION. * AOB = X FHG.
PROOF. * AOB is half a st. * ; also X FHG is half a st. 2.

(66. A right angle is half a straight angle.)

By 98, all straight angles are equal,

. * AOB = 4 FHG.
(93. Halves of equals are equal.)

105. COROLLARY.

From a point on a line, there cannot be more than one perpendicular to that line.

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HYPOTHESIS. X AOA is a perigon; also * DHD is a perigon, CONCLUSION. AOA = X DHD.

PROOF. Any line through the vertex of a perigon divides it into two straight angles. By 98, all straight angles are equal,

.: Perigon at O equals perigon at H.

(92. Doubles of equals are equal.)

107. COROLLARY. From the preceding demonstration, it follows that half a perigon is a straight angle.

THEOREM III.

108. If two lines intersect each other, the vertical angles are equal.

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HYPOTHESIS. AC and BD are two lines intersecting at O.
CONCLUSIONS. AOB = X COD.

* BOC

* DOA, Proof. Because the sum of the angles AOB and BOC is the angle AOC, and by hypothesis A0 and OC are in one line,

:: AOB + X BOC = st. 4.

In the same way,

X BOC + X COD = st. 4,
* AOB + X BOC = X BOC + X. COD.

Take away from these equal sums the common angle BOC, and we have

* AOB = X COD.
(89. If equals be taken from equals, the remainders are equal.)

In the same way we may prove

* BOC = * DOA.

EXERCISES. 1. Two angles are formed at a point on one side of a line. Show that the lines which bisect these angles contain a right angle.

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