XXV. An isosceles triangle is that which has only two sides equal. XXVI. A scalene triangle is that which has three unequal sides. XXVII. A right-angled triangle is that which has a right angle. XXVIII. An obtuse-angled triangle is that which has an obtuse angle. 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. XXXIV. All other four-sided figures besides these are called Trapeziums. XXXV. Parallel straight lines are such as are in the same plane, and which, being produced ever so far both ways, do not meet. POSTULATES. I. 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. AXIOMS. I. Things 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. V. 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. The whole is greater than its part. IX. X. Two straight lines cannot inclose a space. XI. All right angles are equal to one another. 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 continually produced, shall at length meet upon that side on which are the angles which are less than two right angles." 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 (Postulate 3.) the circle BCD; and from the centre B, at the distance BA, describe the circle ACE; and from the point C, in which the circles cut one another, draw (Post. 1.) the straight lines CA, CB, to the points 4, B; ABC shall be an equilateral triangle. C E Because the point A is the centre of the circle BCD, (Definition 15.) 1. AC is equal to AB; and because the point B is the centre of the circle ACE, 2. 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 (Ax. 1.); therefore 3. CA is equal to CB. Wherefore CA, AB, BC, are equal to one another; and therefore 4. The triangle ABC is equilateral; and it is described upon the given straight line AB. Which was to be done. PROP. II.-PROBLEM. From a given point to draw a straight line equal to a given straight line. Let 4 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 (Post. I.) the straight line AB; and upon it describe (I. 1.) the equilateral triangle DAB, and produce (Post. 2.) the straight lines DA, DB to E and F. From the centre B at the distance BC, describe (Post. 3.) the circle CGH, and from the centre D, at the distance DG, describe the circle GKL. AL shall be equal to BC. K H D B G F Because the point B is the centre of the circle CGH, (Def. 15.) 1. BC is equal to BG; and because D is the centre of the circle GLK, 2. DL is equal to DG; 3. and DA, DB, parts of them are equal (Construction); therefore, (Ax. 3.) The remainder AL is equal to the remainder BG. But it has been shown 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 4. 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. PROP. III.-PROBLEM. 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 greater. It is required to cut off from AB, the greater, a part equal to C, the less. From the point 4, draw (I. 2.) the straight line AD equal to C; and from the centre A, and at the distance AD, describe (Post. 3.) the circle DEF: AE shall be equal to C. D E B Because A is the centre of the circle DEF, (Def. 15.) 1. AE is equal to AD; but the straight line C is likewise equal to AD (Constr.); whence AE and Care each of them equal to AD; wherefore (Ax. 1.) 2. The straight line AE is equal to C, and from AB, the greater of two straight lines, a part AE has been cut off equal to 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; they shall likewise have their bases, or third sides equal; and the two triangles shall be equal; and their other angles |