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An isosceles triangle, is that which has only two sides equal.
XV. A circle is a plane figure contained by one line, which is called
the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another:
XVII. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference.
XVIII. A semicircle is the figure contained by a diameter and the part of the circumference cut off by that diameter.
XIX. “A segment of a circle is the figure contained by a straight line, and the circumference it cuts off.”
XX. Rectilineal figures are those which are contained by straight lines.
XXIII. Multilateral figures, or polygons, by more than four straight lines.
XXIV. Of three sided figures, an equilateral triangle is that which has
three equal sides.
A A. Δ
XXIX. An acute-angled triangle, is that which has three acute angles.
XXX. of four-sided figures, a square is that which has all its sides
equal, and all its angles right angles.
XXXI. An oblong, is that which has all its angles right angles, but has not all its sides equal.
XXXII. A rhombus, is that which has all its sides equal, but its angles
are not right angles.
XXXIII. A rhomboid, is that which has its opposite sides equal to one
another, but all its sides are not equal, nor its angles right angles.
which being produced ever so far both ways, do not meet.
I. Let it be granted that a straight line may be drawn from any one point to any other point.
distance from that centre.
The whole is greater than its part.
XII. “If a straight line meet two straight lines, so as to make the two
“ interior angles on the same side of it taken together less “ than two right angles, these straight lines being continual“ ly produced, shall at length meet upon that side on which “ are the angles which are less than iwo right angles. See " the notes on prop: 29. of Book I.”
PROPOSITION I. PROBLEM. To 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.
C From the centre A, at the distance AB, describe (3. Postulate.) the circle BCD, and from the centre B, at the distance BA, describe the fD
E E circle ACE; and from the point C, in which the circles cut one another, draw the straight lines (2. Post.) CA, CB to the points A, B; ABC shall be an equilateral triangle.
Because the point A is the 'centre of the circle BCD, AC is equal (15. Definition.) to AB; and because the point B is the centre of the circle ACE, 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 same are equal to one another ; (1st Axiom.) therefore CA is equal to CB; wherefore CA, AB, BC are equal to one another; and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB. Which was required to be done.
PROP. II. PROB. From 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.
From the point A to B draw (1. Post.) the straight line AB; and up
K on it describe (1. 1.) the equilateral triangle DAB, and produce (2.
H Post.) the straight lines DA, DB, to E and F; from the centre B, at the distance BC, describe (3. Post.) the circle CGH, and from the cen
B tre D, at the distance DG, describe the circle GKL. AL shall be equal