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the circumference, and is such that all straight lines drawn from
A XV A
equal, and all its angles right angles.
are not right angles.
ther, but all its fides are not equal, nor its angles right angles.
which, being produced ever so far both ways, do not meet.
POSTULAT E S.
L any one point to any other point
distance from that center.
A X I O M S.
If equals be added to equals, the wholes are equal.
“ two interior angles on the same side of it taken together less « than two right angles, these straight lines being continually “ produced shall at length meet upon that fide on which are “ the angles which are less than two right angles. See the
notes on Prop. 29. of Book I.”
Book I. PROPOSITION I. PROBLEM. o describe an equilateral triangle upon a given
finite straight line. Let AB be the given straight line, it is required to describe an equilateral triangle upon
it. From the center A, at the distance AB describe a the circle
a. 3d Poftula
late. BCD. and from the center B, at the distance BA describe the circle D A В Е ACE; and from the point C in which the circles cut one another draw the straight lines 6 CA, CB
b. 2d Poft. to the points A, B. ABC shall be an equilateral triangle.
Because the point A is the center of the circle BCD, AC is equal e to AB. and because the point B is the center of the circle ACE, C. 15th De.
finition. BC is equal to BA. but it has been proved that CA is equal to AB; therefore CA, CB are each of them equal to AB. but things which are equal to the fame are equal to one another d; therefore d. ift Axia CA is equal to CB. wherefore CA, AB, BC are equal to one another. and the triangle ABC is therefore equilateral, and it is defcribed upon the given straight line AB. Which was required to be done.
a. I. Poft.
PRO P. II. P R O B,
to a given straight line.
From the point A to B draw a the Straight line AB; and upon it de
K fcribe the equilateral triangle DAB,
H and produce the straight lines DA,
c. 2. Poft, DB to E and F; from the center B, at the distance BC describe d the
d. 3. Poft. circle CGH, and from the center D,
B at the distance DG described the
E circle GKL. AL Thall be equal to BC.
b. I. I.