Part. Enun.-Let the three 48 of the A ABC be bi. sected by the st. lines AD, BD, CD (Prop. IX.); produce AD to E (Post. 2), and from draw DF I to BC (Prop. XII.); then the 2 BDE = L CDF. DEMONST.-1. Because the three /s of A ABC = two rt. 18 (Prop. XXXII.), • . the $ DBA + DAB + DCF = a rt. (Const.) 48 DCF + CDF (Prop. XXXII.) ; by subtraction, 28 DBA + DAB = 2 CDF (Ax. 3). 2. But the exterior / BDE of A ABD 48 DBA + DAB (Prop. XXXII.); B .. the L BDE = L CDF (Ax. 1). Wherefore, if the three angles &c.-Q. E. D. E F PROP. BB. THEOR. F Gen. Enun.-If the sides of an equilateral and equian. gular pentagon, or five-sided figure, be produced to meet, the angles formed by these lines are together equal to two right angles ; and if the sides of an equilateral and equiangular hexagon, or six-sided figure, be produced to meet, the angles so formed are equal to four right angles. Part. Enun.-1. Let ABCDE be an equilateral and equi. angular pentagon; produce its sides to meet in F, G, H, K, L; then the Ls at these pts. together two rt. ZS. DEMONST.–For the L ABF (the exterior of A LBH) s at h and , and the L BAF (the exterior L of A GAK) Sat G and K (Prop. XXXII.); .., by addition, 28 ABF + LS G, H, K, L; add to each the < SAFB; .. the L at F, G, H, K, L = the three /s of ABF 2 rt. Ls. (Prop. XXXII.) B в E K D н BAF = N M Part. ENUN.-2. Let ABCDEF be an equilateral and equiangular hexagon; produce its sides to meet in G, H, K, L, M,N; the Zs at these pts. = four rt. 45. Demonst.–For they are the <s of the two AS GMK, HLN: and the Ls of any A = two rt. { $ (Prop. XXXII.); the six { $ at G, H, K, L, M, N = four H rt. L$. Wherefore, if the sides &c. Q. E. D. B E .. K. PROP. XXXIII. THEOR. A B = and Gen. Enun.—The straight lines, which join the extremities of two equal and parallel straight lines towards the same parts, are themselves also equal and parallel. Part. Enun.—Let AB, CD be = and || st. lines, and let them be joined towards the same parts by the st. lines AC, BD; then AC is ll to . Const.-Join BC. DEMONST.–1. Because AB = CD, and BC is common to the 2 AS ABC, BCD, .. the two sides AB, BC = DC, CB, each to each : and because AB is || to cd, the alternate z ABC = alternate 2 BCD (Prop. XXIX.); .. the base AC = the base bd. (Prop. IV.) 2. Also the 2 ACB = 2 CBD. Hence, because the st. line Bc meets the 2 st. lines AC, BD, and makes the alternate ZS ACB, CBD = to one another, ... Ac is || to BD. And it has been shewn that AC = BD. PROP. XXXIV. THEOR. А. B в are GEN. ENUN.—The opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them into two equal parts. N. B. A parallelogram is a four-sided figure, of which the opposite sides are parallel : and the diameter is the straight line joining two of its opposite angles. PART. ENUN.—Let ABCD be a om, and BC its diameter; then the opposite sides and 28 = to one another, and the diameter bc bisects it. DEMONST.-1. Because AB is || to CD, and BC meets them, .. the alternate z ABC = alternate BCD (Prop. XXIX.): and because AC is 11 to BD, and bc meets them, .. the alternate 2 ACB = alternate < CBD (Prop. XXIX.); . . in the A$ ABC, CBD, the < $ ABC, BCA = Z $ BCD, CBD, each to each, and the side BC, adjacent to these = 2$, is common; .. the remaining sides are =, each to each, and the third to the third 2 (Prop. XXVI.): i.e. AB = CD, AC = BD, and the Z BAC = 2 BDC. 2. And because the 2 ABC = Z BCD, and ms are = the Z CBD = L ACB, ... the whole 2 ABD = the whole 2 ACD: therefore the opposite sides and Zs of 3. Also because AB = CD, and bc is common to the As ABC, BCD, .. the two AB, BC = DC, CB, each to each, and the included – ABC = included < BCD; .. the A ABC (Prop. IV.): i.e. the diam. BC bisects the ABCD. Wherefore the opposite sides &c.—Q. E. D. A few important Propositions are deducible from this Theorem, which, after first demonstrating its converse, so far as it is convertible, it may be proper to subjoin. ABCD m PROP. CC. THEOR. GEN. ENUN.-If the opposite sides, or the opposite angles, of a quadrilateral figure be equal, the figure is a parallelogram. Part. Enun.-Let ABCD be a quadrilateral figure, of which the opposite sides, or opposite 28, are = ; then ABCD is a om. DEMONST.–First, let the opposite sides be =. Join CB. А B В 1. Then, because AB = CD, and Bc common to A$ ABC, DBC, . , the two AB, BC = DC, CB, each to each, and the base AC = base BD; .. the L ABC | BCD (Prop. VIII.), and they are alternate 4S; .. AB is il to cd. (Prop. XXVII.) 2. Also the L ACB = and they are alternate 48; .. also ac is | to BD, and .. ABCD is a om. (Def.) 3. Again, let the opposite 2s be =. Now the Z interior of the quadrilateral ABCD 4 right (Prop. XXXII. Cor. 1); also the ZS BAC + ACD LS ABD + BDC D LCBD, (Hyp.); .. the 48 BAC + ACD = 2 rt. L, and ... Ab is ll to cd. (Prop. XXVIII.) 4. In the like manner, it may be shewn that ac is | to BD; . . ABCD is a om. Wherefore, if the opposite sides &c.-Q. E. D. Cor.—Hence it appears, from the definitions of the square, and the oblong, the rhombus, and the rhomboid, that these figures are ams. That the latter part of the 34th Proposition is not convertible, may be inferred from the following : PROP. DD. THEOR. Gen. Enun.-A trapezium, of which the adjacent sides are equal, is bisected by the diameter. Part. Enun.—Let ABCD be a trapezium, of which the sides AB = bd, and ca = cd; then the diam. cb bisects it. DEMONST.-Because AB = BD, Α. base cd; .. the A ABC A DBC (Prop. VIII. Obs.); and the trapezium is bisected by diam. Bc. Wherefore a trapezium &c.-Q. E. D. E D PROP. EE. THEOR. GEN. ENUN.—The diameters of a parallelogram bisect each other; and, vice versá, if the diameters of a quadrilateral figure bisect each other, the figure is a parallelogram. Part. Enun.-Let ABCD be a parallelogram; its diams. AD, BC bisect each other in E, and vice versa. DEMONST.-1. Because ac is Il to BD, and cB meets them, ... the alternate | ACE = alternate / EBD (Prop. XXIX.); and because ac is || to BD, and AD meets them, the alternate L CAE= alternate / EDB (Prop. A B в E |