meets the parallels KM, FG, the alternate angles MHG, HGF are equal ©: Add to each of these the angle HGL: Therefore the angles MHG, HGL are equal to the angles HGF, HGL: с But the angles MHG, HGL are equal to two right angles; wherefore also HGF, HGL are equal to two right angles, and FG is therefore in the same straight line with GL; and because KF is parallel to HG, and HG to ML; KF is parallel to ML; and KM, FL are parallels; wherefore KFLM is a parallelogram; and because the triangle ABD is equal to the parallelogram HF and the triangle DBC to the parallelogram GM; the whole figure ABCD is equal to the whole KFLM; COR. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. [The drawing the figure of this Proposition may advantageously be omitted at the first reading. Indeed the process of drawing it being a mere repetition of the processes employed in XLII. and XLIV. may be omitted altogether, if the figures in XLII. and XLIV. have been once carefully drawn.] BOOK I. a xi. 1. b iii. 1. PROP. XLVI. PROB. To describe a square upon a given straight line. Let AB be the given straight line; it is required to describe a square upon AB. From the point A draw a AC at right angles to AB; and make AD equal to AB, ⚫ xxxi. 1. and through the point D draw DE parallel to AB, and through B draw BE parallel to AD; Therefore ADEB is a parallelogram: a xxxiv. 1. whence AB is equal to DE, and AD to BE: But BA is equal to AD; therefore the four straight lines are equal to one another, and the parallelogram ADEB is equilateral, likewise all its angles are right angles; because the straight line AD meet ing the parallels AB, DE, D A • xxix. 1. the angles BAD, ADE are equal to two right angles: but BAD is a right angle; therefore also ADE is a right angle; B but the opposite angles of parallelograms are equala; wherefore the figure ADEB is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB. COR. Hence every parallelogram that has one right angle BOOK I. has all its angles right angles. [As it is convenient to erect a perpendicular at the end of a line without producing it, use the following method: Take any point B on the line required. On AB con struct an equilateral triangle AOB. Produce BO to P cutting off OP equal to OB. PA shall be perpendicular to АВ. The proof that it is so is as follows: the angle P is equal to the angle OAP, and the angle B is equal to the angle OAB. Add these equals, and the angles P and B together are equal to the angles PAO and OAB, or to Practise constructing squares, as follows: B Р BOOK I. Upon AB construct a square, and upon each of its sides construct squares, and upon one of the exterior lines of the figure construct another square. Cut the figure out in cardboard, and let the squares be moveable on the sides of the first constructed square as hinges, and the last drawn square upon the side on which it was constructed: turn them till the points O, P, Q coincide. The solid figure formed is called a cube, and is the second of the five regular solids.] . xlvi. 1. PROP. XLVII. THEOR. In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle. Let ABC be a right-angled triangle having the right angle BAC; the square described upon the side BC is equal to a On BC describe the square BDEC, and on BA, AC the squares GB, HC; b xxxi. 1. and through A draw AL parallel to BD, or CE, and join AD, FC; for the same reason, AB and AH are in the same straight BOOK I. line; and because the angle DBC is equal to the angle FBA, each of them being a right angle, add to each the angle ABC, and the whole angle DBA is equal to the whole FBC, and because the two sides AB, BD are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is is doubles of the tri- because they are upon the same base BD, and between the same pa rallels, BD, AI; and the square GB is double of the triangle FBC, because these are also upon the same base FB, and between the same parallels FB, GC. But the doubles of equals are equal to one another: Therefore the parallelogram BL is equal to the square GB: And, in the same manner, F B e ⚫ 2 Ax. A f iv. 1. C xli. 1. by joining AE, BK it is D L E demonstrated, K h 6 Ax. |