PROPOSITION 44. PROBLEM. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. 1. Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle: it is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to D. 2. Make the parallelogram BEFG equal to the tri angle C, and having the angle EBG equal to the angle D, so that BE may be in the same straight line with AB; [I. 42.] produce FG to H; through A draw AH parallel to BG or EF, [I. 31.) and join HB. Then because AH is parallel to FE, HB is not 3. parallel to FE. [Ax. 12.] Therefore HB and FE will meet if produced ; let them meet in K. 2a. Through K draw KL parallel to EA or FH; [I. 31.] and produce HĂ, GB to the points L, M. Then HLKF is a parallelogram, of which the 3a. diameter is HK; and AG, ME are parallelograms about HK; and LB, BF are the complements. Therefore LB is equal to BF. [I. 43.] But BF is equal to the triangle C. [Const.] Therefore LB is equal to the triangle C. [Ax. 1.] And because the angle GBE is equal to the angle ABM, [I. 15.] and likewise to the angle D; [Const.] the angle ABM is equal to the angle D. [Ax. 1.] Therefore to the given straight line AB the parallelogram LB is applied, equal to the triangle C, and having the angle ABM equal to the anglo D. Q.E.F. Ex. On a given straight line construct a right-angled triangle equal in area to a given triangle. PROPOSITION 45. PROBLEM. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and 1. E the given rectilineal angle: it is required to describe a parallelogram equal to ABCD, and having an angle equal to E. А. L Join DB, and describe the paralellogram FH 2. equal to the triangle ADB, and having the angle FKH equal to the angle E; [I. 42] and to the straight line GH apply the parallelogram GM equal to the triangle DBC, and having the angle GHM equal to the angle E. [I. 44.] The figure FKML shall be the parallelogram required. Because the angle E is equal to each of the 3. angles FKH, GHM, [Const.] the angle FKH is equal to the angle GHM. [Ax. 1.] Add to each of these equals the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM. [Ax. 2.] But FKH, KHG are equal to two right angles ; [I. 29] therefore also KHG, GHM are equal to two right angles. And because at the point H in the straight line GH, the two straight lines KH, HM, on the opposite sides of it, make the adjacent angles together equal to two right angles, KH is in the same straight line with HM. [I. 14:] And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal. [I. 29.] Add to each of these equals the angle HGL; therefore the angles MHG, HGL, are equal to the angles HGF, HGL. [Ax. 2.] But MHG, HGL, are equal to two right angles ; [I. 29] therefore also HGF, HGL are equal to two right angles. Therefore FG is in the samé straight line with GL. [I. 14.] And because KF is parallel to HG, and HG to ML, [Const.] KF is parallel to ML; [I. 30] and KM, FL are parallels ; Const.] therefore KPLM is a parallelogram. [Def.] And because the triangle ABD is equal to the parallelogram HF, [Const.) and the triangle DBC to the parallelogram GM; (Const.] the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. [-Ax. 2.] Therefore, the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, and having the angle FKM equal to the given angle E. Q.E.F. COROLLARY. 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 ; namely, by applying to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle; and so on. [I. 44.] Ex. Show how to make a triangle equal in area to a given quadri lateral. PROPOSITION 46. PROBLEM. 2. To describe a square on a given straight line. Let AB be the given straight line: it is required 1. to describe a square on AB. AC at right angles to AB; For ADEB is by con3. struction a parallelogram ; therefore AB is equal to DE, and AD to BE. [I. 34.] But AB B is equal to AD. [Const.] There- A fore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral. [Ax. 1.] Likewise all its angles are right angles. For since the straight line AD meets the parallels AB, DE, the angles BAD, ADE are together equal to two right angles ; [I. 29.) but BAD is a right angle ; [Const.] therefore also ADE is a right angle. [Ax. 3.] But the opposite angles of parallelograms are equal. [I. 34.] Therefore each of the opposite angles ABĒ, BED is a right angle. [Ax. 1.] Therefore the figure ADEB is rectangular; and it has been proved to be equilateral. Therefore it is a square. [Def. 30.] And it is described on the given straight line AB. Therefore a square has been described, &c. COROLLARY. From the demonstration it is manifest that every parallelogram which has one right angle has all its angles right angles. Ex. Construct a square having given the length of its diagonal. Q.E.F. Н F PROPOSITION 47. THEOREM. In any right-angled triangle, the square which is described on the side subtending the right angle is equal to the squares described on the sides which contain the right angle. Let ABC be a right-angled triangle, having the 1. right angle BAC: the square described on the side BC shall be equal to the squares described on the sides BA, AC. On BC de2. scribe the square BDEC, and on BA, AC describe the squares GB, HC; [I. 46.] through A draw AL parallel to K BD or CE; [I. 31.] and join AD, FC. Then, be3. the angle BAC is a right angle, [Hyp.] and that the angle BAG is also a right angle, [Def. 30.] the two E straight lines AC, AG, on the opposite sides of AB, make with it at the point A, the adjacent angles equal to two right angles; therefore CA is in the same straight line with AG. [I. 14.] For the same reason, AB and AH are in the same straight line. Now the angle DBC is equal to the angle FBA, for each of them is a right angle. [Ax. 11.] Add to each the angle ABC. Therefore the whole angle DBA is equal to the whole angle FBC. Ax. 2.] And because the two sides AB, BD are equal to the two sides FB, BC, each to each ; [Def. 30.] and the angle DBA is equal to the angle FBC; therefore the base AD is equal cause |