XI. 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. XII. And this point is called the centre of the circle. XIII. A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. a XIV. A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter. XV. XVI. XVII. XVIII. Multilateral figures, or polygons, by more than four straight lines. XIX. Of three sided figures, an equilateral triangle is that which has three equal sides. XX. AAN XXI. XXII. XXIII. XXIV. XXV. and all its angles right angles. XXVI. An oblong, is that which has all its angles right angles, but has not all its sides equal. XXVII. A rhombus, is that which has all its sides equal, but its angles are not right angles. XXVIII. 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. XXIX. XXX. being produced ever so far both ways, do not meet. POSTULATES. Let it be granted that a straight line may be drawn from any one point to any other point. II. That a terminated straight line may be produced to any length in a straight line. III. And that a circle may be described from any centre, at any distance from that centre. a AXIOMS. I, II. III. IV. If equals be taken from unequals, the remainders are unequal. VI. Things which are doubles of the same thing, are equal to one another. VII. Things which are halves of the same thing, 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. X. XI. “ Two straight lines which intersect one another, cannot be both “parallel to the same straight line.'' 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. 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 circle ACE ; and from the point C, in D B E which the circles cut one another, draw the straight lines (1. Post.) CA, CB to the points À, B; ABC is an equilateral triangle. Because the point A is the centre of the circle BCD, AC is equal (11. Definition) to AB; and because the point B is the centre of the circle ACE, BC is equal to AB : But it has been proved that CA is equal to AB; therefore CA, CB are each of them equal to AB; now things which are equal to the same are equal to one another (1. Axiom); therefore CA is equal to CB; wherefore CA, AB, ČB 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 re ; quired 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 upon it describe (1. 1.) the equilateral triangle DAB, and produce (2. Post.) the straight lines DA, H BD, to È and F; from the centre B, at the distance BC, describe (3. Post.) the D circle CGH, and from the centre D, at the distance DG, describe the circle GKL. AL is equal to BC. Because the point B is the centre of the circle CGH, BC is equal (11. Def.) to BG; and because D is the centre of the e circle GKL, DL is equal to DG, and F DB, parts of them, are equal; therefore A the remainder AL is equal to the remainder (3. Ax.) 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 lině BC. Which was to be done. PROP. III. PROB. From the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the D greater. It is required to cut off from AB, the greater, a part equal to C, the less. From the point A draw(2. 1.) А. E the straight line AD equal to C; B and from the centre A, and at the distance AD, describe (3. Post.) the circle DEF ; and because A is the centre of the circle DEF, AE is equal to AD; but the straight line C is likewise equal to AD; whence AE and C are each of them equal to AD; wherefore the straight line AE is equal to (1. Ax.) C, and from AB the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done. PROP. IV. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each ; and have likewise the angles contained by those sides equal to one another, their bases, or third sides, shall be equal ; and the areas of the triangles shall be equal ; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite*. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF ; and let the A D angle BAC be also equal to the angle EDF : then shall the base BC be equal to the base EF ; and the triangle ABC to the triangle DEF ; and the other angles, to which the equal sides are op B C * The three conclusions in this enunciation are more briefly expressed by saying, that the triangles every way equal. |