Let ABCD be the given square. Draw the diagonals AC, BD, intersecting in E. Through A draw AF parallel to BE; and through B draw BF parallel to AE. Then AEBF is the square required. For, by construction it is a parallelogram; and one of its angles, viz. AEB, is a right angle, since the triangles AEB, AED are equal in all respects; also AE=EB; .. all its angles are right angles, (42 Cor. 2) and all its sides equal E B (40). It is also equal to the two triangles AEB, AED, which are together half the square ABCD. 122. PROP. XXV. To construct a square which shall be any multiple of a given square. F K G C Let ABCD be the given square; produce AB, AD, indefinitely towards B and D, join BD; in AB produced take AE = BD; and in AD produced AF =BD. Draw FG parallel to AB, and EG parallel to AD. Then AEGF is a square, and it is double of the square ABCD. Again, join BF; take AH = BF, and AI=BF; complete the square D A AHKI, and it is three times the on for succeeding multiples. For, AEGF square of AE, Also, AHKI: = = = = square square of BD, square of AB+ square of AD (43), = twice the square of AB, = twice ABCD. = = square of AH, square of AB + square of AF, = square of AB+twice square of AB, = three times the square of AB; = and so on for succeeding multiples. 1 123. PROP. XXVI. To construct a rectangle with sides equal to given straight lines. Since the opposite sides of every rectangle are equal to one another (13), two straight lines only are needed to be given in this case. D Draw the straight line AB equal to one of the given lines, and from the extreme points in it draw AC and BD at right angles to AB, making each of them equal to the other given line; and join CD. Then ACDB is obviously the rectangle required. A The correctness of the work may be tested, as in the square, by measuring the diagonals AD, BC, which ought to be equal to one another. 124. PROP. XXVII. To construct a rectangle which shall be equal to a given parallelogram, DE C F Let ABCD be the given parallelogram; produce the side DC; and from the extreme points of the base AB, draw AE, BF perpendicular to DC, and DC produced. Then ABFE is the rectangle required. A B For ABFE is a parallelogram, and ABFE, ABCD are upon the same base AB, and between the same parallels (41). 125. PROP. XXVIII. To construct a rectangle which shall be equal to a given triangle. D G E Let ABC be the given triangle; through C draw DCE parallel to AB; and through A and B draw AD, BE at right angles to AB; bisect AB in F, and through F draw FG at right angles to AB; AFGD, or BFGE, is the rectangle required. For the triangle ABC, and the parallelogram ABED, B are on the same base AB, and between the same parallels, .. the triangle is equal to half the parallelogram (41 Cor. 2). Also, since AFBF, the parallelograms ADGF, BEGF are equal to one another (41 Cor. 1), that is, each of them is half of ABED; :. each of them is equal to the triangle ABC. 126. PROP. XXIX. To construct a square which shall be equal to a given rectangle. Let ABCD be the given rectangle; produce one side AB to E, making BE=BC, the other side of the rectangle: bisect AE in F; with centre F, and radius AF, describe a semi-circle AGE, producing CB to meet the circumference in G. The A of BG is the square square required. For, joining AG, EG, D AGE is a right angle, T B E being the 'angle in a semi-circle', .. BG is a 'mean proportional' between AB, and BE, that is, between AB and BC (72 Cor. 2); and .. the rectangle AB, BC = the square of BG (74). 127. PROP. XXX. To construct a lozenge with its side equal to a given straight line. Let A be the given straight line; take any straight line BC, less than twice A; with centres B and C, and radius equal to A, describe intersecting arcs at D and E, on opposite sides of BC. Join BD, BE, CD, CE; and BDCE is the lozenge required. The proof is obvious; and there will be various A B C forms, all satisfying the conditions of the question, according to the assumed magnitude of BC. It is to be remembered, however, that the case is excluded, in which the diagonals BC, DE are equal, because then the figure BDCE will be a square, and not a lozenge. It may also be worth remembering, that besides the common property of equal sides, the square and lozenge have another property in common, viz. that the diagonals are in every case at right angles to one another, as may easily be proved. 128. PROP. XXXI. To construct a lozenge with given diagonals, that is, with diagonals equal respectively to two given straight lines. D Take AB equal to one of the given lines; bisect AB in C; from C draw CD, and CE, on opposite sides of AB, and at right angles to AB; and make CDCE= half of the other given straight line. Join AD, AE, BD, BE, and ADBE is the lozenge required. That it is a lozenge will easily be shewn by (24); and the diagonals are equal to the given lines by construction. 129. PROP. XXXII. A E B To construct a lozenge which shall be double of a given lozenge, and have one diagonal the same. F Let ABCD be the given lozenge, and AC the given diagonal. Join BD, meeting AC in E; and produce it both ways to F and G, making DF = DE = BG. Join AF, AG, CF, CG, and AFCG is the lozenge required. For, since the triangles DEA, FDA are upon equal bases DE, DF, and between the same parallels', they are equal to one another (41 Cor. 2); the triangle AEF is double of the triangle AED. But AEF is one-fourth of the lozenge AFCG; and AED is one fourth of the given lozenge; .. AFCG is double of ABCD. E B 130. PROP. XXXIII. To explain the T square, and the Drawing-Board. D (1) The Drawing-Board is a smooth board, as ABCD, very accurately rectangular, with its edges quite smooth. Upon this board drawing-paper is usually fixed, and evenly stretched, by means of glue or paste applied to a small strip of the paper all round, which it is not intended to retain in the drawing. A B E F (2) The T square is an instrument consisting of two parts, one called the stock, as EF, into which the other, called the blade, is fixed, near the middle of it, at right angles to EF. The blade is a thin flat-ruler, and the stock is somewhat thicker; so that there is a projecting edge of the stock, which enables the draughtsman to slide the square backwards or forwards along the drawingboard, keeping EF in close contact with AB, while the blade continues to lie flat, in every position of the square, upon the drawing-paper. Thus, the blade being a ruler which is always at right angles to AB, it is evident that any number of parallel lines may be drawn along it on the paper which is fixed upon the Drawing-Board. And, again, by removing the square to one of the adjacent sides of the board, as AD, it is also evident that any number of lines may be drawn at pleasure at right angles to the former, and parallel to one another. It is scarcely necessary to point out the advantage of the above combination to architectural draughtsmen, or to any others, who are required in the same drawing to trace a series of lines parallel to each other. Of course, for accurate work both the Drawing-Board and the Square must be accurately constructed. The former may be tested by measuring its opposite sides, and its diagonals. For, in a true rectangle ABCD, AB= CD, AD=BC, and AC=BD, the whole three conditions must be satisfied. Also the T square may be tested in the manner pointed out for the ordinary square in (96). |