XXIX.); .. in A$ AEC, BED, the LS ACE, CAE EBD, BDE, each to each, and, adjacent to each, AC = BD; . . AE = ED, and be = EC (Prop. XXVI.): therefore the diams. AD, BC, mutually bisect each other. 2. But, let the diams. AD, BC bisect each other in E; then, since, in the AS AEC, BED, the two AE, EC = the two DE, EB, each to each, and the included vertical L AEC included vertical Z BED (Prop. XV.); .. the base AC = base BD, and the L ACE = L EBD (Prop. IV.): 1. Ac is U to BD. In like manner, AB is || to cD; ABCD is om. Wherefore the diameter &c.-Q. E. D. PROP. FF. PROB. E A F Gen. Exun.-To bisect a parallelogram by a straight line drawn from a given point in one of its sides. Part. Enun.-Let ABCD be a om, E a gn. pt. in the side AB; then it is required to bisect ABCD by a line drawn through E. Const.-Draw the diam. BC, and bisect it in F (Prop. X.); join EF, and produce it to G (Post. 1 and 2); then the om ABCD is bisected by Eg. DEMONST.--Because AB is || D to cd, and bc meets them; .. the alternate L EBF =- alternate / FCG (Prop. XXIX.), and the vertical 4 EBF = vertical / GCF (Prop. XV.), also BF = Fc (Const.); .. the A EFB = A GCF. (Prop. XXVI.) 2. But A ABC = ABCD (Prop. XXXIV.); .., by subtraction, trapezium AEFC = trapezium BFGD (Ax. 3). 3. Hence, by addition, the trapezium AEGC = pezium EBDG (Ax. 1). Wherefore the Om ABCD has been bisected by the line EG drawn through the gn. pt. E.-Q. E. F. G PROP. GG. THEOR. C Gen. ENUN.-The diameters of rectangular parallelograms are equal to one another; and in any other parallelogram the diameter which joins the acute angles is greater than that which joins the obtuse. PART. ENUN.-Case 1. Let ABCD be a rectangular om, as a square or oblong; A then the diam. AD = diam. CB. DEMONST.-Since, in the AS ABD, BAC, the two AB, BD = BA, AC, each to each, and the rt. L ABD rt. Z BAC (Def. 10), · the base AD = the base cs. (Prop. IV.) Part. ENUN.–Case 2. Let Abcd not be rectangular, as a rhombus or rhomboid, of which the opposite / S ABD, ACD are acute, and CAB, CDB, obtuse; then BC is > than AD. DEMONST. For, as before, the two AB, BD = BA, AC, each to each; but the / Bac is > than / ABD, .. the base BC is > the base ad. (Prop. XXIV.) Wherefore the diameters &c.-Q. E. D. CoR.-Since the diams. of a am bisect each other (Prop. EE), if the hypothenuse of a rt. 2d A be bisected, the st. line drawn from the rt. Z to the pt. of bisection is to half the hypothenuse. B D D PROP. XXXV. THEOR. Gen. ENUN.-Parallelograins upon the same base, and between the same parallels, are equal to one another. ms PART. ENUN.- Let the ABCD, EBCF (Fig. 2 and 3) be upon the same base bc, and between the same || S AF, BC; then the om EBCF. Fig. 3. ABCD = m А ת FA E A ED F B S DEMONST.–1. If the sides AD, DF of the ABCD, DBCF (Fig. 1), opposite to the base BC, terminate in the same pt. D, it is plain that each of them, being bisected by its diameter (Prop. XXXIV.), = 2 A DBC; .. the om ABCD = Om DBCF (Ax. 6). 2. But if the sides AD, EF, opposite to the base sc (Figs. 2 and 3), be not terminated in the same pt.; then, by the property of a om AD = BC = EF. (Prop. XXXIV.) 3. To or from each of these equals add or subtract DE; .. AD + DE (or AE) = EF + DE (or DF). Also AB = 'DC. 4. : . in the two A$ EAB, FDC, the two EA, AB = FD, DC, each to each, and the interior < EAB = exterior Z FDC (Prop. XXIX.); EAB = A FDC. (Prop. IV.) 5. Take each of these as from the trapezium ABCF, and the remainders will be = (Ax. 3): i. e. the o ABCD = OM EBCF. Wherefore or upon the same base, &c.Q. E. D. m The equality of ams is that of their areas; and it appears from this Proposition that any oblique m is equal to a rectangular om, or rectangle, upon the same base and between the same || $, i. e. of the same base and altitude. Now it is known that the area of a rectangle the base x its altitude; and .. generally the area of any om: base x the I altitude. The converse of the Proposition is also true. the PROP. HH. THEOR. ms A D F G B Gen. Enun.-Equal parallelograms, upon the same base, are between the same parallels. PART. ENUN.—Let the ABCD, EBCF be upon the same base BC; then they are between the same || 5: i.e. EF is in the same straight line with AD. DEMONST.-1. If not, let Eg be in the same st. line with AD; . . EG is ll to BC, and EBCG is a am. (Prop. XXXIV.) 2. Now, because the ams ABCD, EBCG are upon the same base, and between the same || , they are = to one another. 3. But the am ABCD = OM EBCF; .. the OM EBCG = OM EBCF, or the < >, which is impossible. 4. .. EG is not in the same st. line with AD. 5. In like manner, it may be proved that no other line but Er is in the same st. line with AD; .. AD, EF are in the same st. line, and ... || with Bc. (Prop. XXXIV.) Wherefore oms &c.-Q. E. D. PROP. XXXVI. THEOR. Gen. ENUN.-Parallelograms upon equal bases, and between the same parallels, are equal to one another. ms A D E same S B PART. ENUN.--Let ABCD, EFGH be upon equal bases BC, FG, and between the ll AH, BG; then the am ABCD = OM EFGH. Const.–Join BE, сн. DEMONST.-1. Because BC = FG, and FG = EH, .'. BC = EH (Ax. 1). 2. But bc, Eh are ll, and joined towards the same parts by BE, CH; .'. BE and ch are also = and || (Prop. XXXI.); and ... EBCH is a om. (Prop. XXXIV. Def.) 3. Now, because the oms ABCD, EBCH are upon the same base BC, and between the same || S AH, BC, .. ABCD = OM EBCH. (Prop. XXXV.) 4. For the same reason, o EFGH = Om EBCH;.'. om EFGH (Ax. 1). Wherefore ms &c.—Q. E. D. The converse of this Proposition may be demonstrated in the same manner as that of the foregoing; and it is left to the student to shew, that equal oms, upon bases in the same straight line, are between the same || $. (See also Prop. XL.) The following are deductions : m m m ABCD = PROP. KK. THEOR. GEN. ENUN.-If two sides of a trapezium are parallel, its area is equal to half that of a parallelogram, between the same parallels, whose base is equal to the two parallel sides taken together. |