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I.
ET it be granted that a straight line may be drawn from
any one point to any other point.

That a terminated straight line may be produced to any length
in a flraight line.
III.

And that a circle may be described from any centre, at any distance from that centre.

A X I O M S.
I

HINGS which are equal to the same are equal to one another. - II. * If equals be added to equals, the wholes are equal. III. If equals be taken from equals, the remainders are equal. IV.

If equals be added to unequals, the wholes are unequal.
If equals be taken from unequals, the remainders are unequal.
VI

Things which are double of the same, are equal to one another.
VII.

Things which are halves of the same, are equal to one another.
VIII.

Magnitudes which coincide with one another, that is, which exactly fill the same space, are equal to one another. IX.

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- --- // 2 × O F E U C L I D. 15 PROPOSITION I. PROB L E M. Book I. o describe an equilateral triangle upon a given fi-\-ented I nite straight line.

Let AB be the given straight line; it is required to describe an equilateral triangle upon it.

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shalí be an equilateral triangle.
Because the point A is the centre of the circle BCD, AC is

equal “to AB3 and because the point B is the centre of the . 13th De.

circle ACE, BC is equal to BA: But it has been proved that CA finition.

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AB; but things which are equal to the same are equal to one
another d, therefore CA is equal to CB; wherefore CA, AB, d. 1st Axis
BC are equal to one another; and the triangle ABC is there. "
fore equilateral, and it is described upon the given straight line -
AB. Which was required to be done.

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ROM a given point to draw a straight line equal to
a given straight line.

Let A be the given point, and BC the given straight line; it is required to draw from the point A a straight line equal to BC.

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therefore the remainder AL is equal to the remainder f BG : But it has been shewn, that BC is equal to BG; wherefore AL and BC are each of them equal to BG; and things that are equal to the same are equal to one another; therefore the straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was to be done.

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ROM the greater of two given straight lines to cut off a part equal to the less.

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Le ABC, DEF be two triangles which have the two fides,
AB, AC equal to the two sides DE, DF, each to each, Xī

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| produced, the angles upon the other fide of the base shall be equal.

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