To divide a Trapezium into two equal Parts, by a Right Line, drawn from a Point given in any Side. ABCD is the Trapezium given, and E the given Point. 3 7 2 Through the Angle B, draw BF, parallel to AD. — 5. Bifect A D and BF, in G and H; and draw GH and HC. Then the Pentagon ABCHG is equal to the Trap. AHCD. DEM. For (having drawn BG and GF) the Triangle ABG is equal to GFD, the Triangle BGH is equal to HGF, and BCH is equal to HCF. P. 18. 1. 2dly. In Fig. 2. let ABCHG be equal GHCD, as before. Draw GC, and HI parallel to GC, cuting BC in I, and draw Gl. The Right Line GI bifects the given Trapezium. DEM. For, because HI is parallel to GC, the Triangle GIC Again. Draw EI, and GK parallel to E I, cuting BC in K, and draw E K. I fay, that EK divides the Trapezium ABCD equally. DEM. DEM. For first, the Trap. GHCD was proved half ABCD; fecondly; GICD was proved equal to GHCD; and, EKCD is equal to GICD. For the Triangle EKI is equal to EGI. P. 18. 1. Therefore, the Trap. ABKE is equal to EKCD. Now, if E had been given near the Angle A, at the leffer end of the Trapezium, the operation would require more labour. Let Fig. 3. be fuppofed the fame Figure divided, by the Line GI, into two equal parts, from the middle Point G (as by the fecond operation) which is neceffary to be first done, in all cafes. Join the given Point, E, and I, as before; draw GF, parallel to EI, cuting CD in F, and draw EF and IF. Then, the Pent. ABIFE is equal to the Tri. EFD +FIC. For, because GF is parallel to EI, the Tri. IFE-EGI. But, although the Pentagon ABIFE is equal to half the Trap. ABCD, it is not equally divided by one Right Line. Therefore, draw IH parallel to CD, cuting EF in H, and draw HC. Then, because IH is parallel to CF, the Triangle ICH is equal to IFH. Wh. the Pent. ABCHEABIFE 18. I. the Trap. EHCD. Laftly; join EC, and draw HK parallel to EC, cuting CF in K; and draw EK; which Line divides the Trapezium ABCD into two equal parts, as required. For, fince HK is parallel to EC the Tri. EKC EHC; wherefore, EKC added to ECBA EHC added to ECBA. But, ABCHE was equal to ABIFE, equal to ABIG; which, was equal to half the Trapezium ABCD. X Y Z E To find the Side of a Square, equal to any Number of Squares. Let X, Y and Z be the Sides of three given Squares. B It is required to find the proportion of a Line, on which if a Square be conftructed, it fhall be equal in Area to all the three. Pr. 10. Make a Right Angle ABD. From the Angle B, take BA and B C, equal to any two of the given Lines, respectively, as X and Y, and join AC. The Square of AC is equal to the two Squares of AB and BC, or X and Y. Again. Make BD equal AC, and BE equal Z; Draw ED, which is the Line required. P. 20. 1. DEM. For, ED square the two Squares of EB & BD. i. e. ED square the three Squares, of X, Y and Z.--20.1. APPL. By this useful Problem, Quantities may be increased in any Proportion at pleasure. Alfo, by means of this Problem, and Prop. 20. 1. Carpenters form a Right Angle, in framing Timber, &c. For, having made AB equal 3 feet, and BC equal four; then, if AC measures 5 feet ABC is a Right Angle. N. B. The Numbers 3, 4 and 5 being multiplied, feparately, by any one Number, at pleasure, produce the fame effect. e. g. If AB be made 6 or 9 feet, and BC equal 8 or 12 feet, then will AC be equal to 10 or 15 feet, if the Angle ABC be a right one. For, the Square of 9 is 81; and the Square of 12 is 144, added to 81 is 225; equal to the Square of 15.-20. 1. El. When When the Timbers are long, it is neceffary, in placing them. at right angles, to apply greater meafures; which, will give the Right Angle with greater accuracy. COR. Hence, a Perpendicular may be drawn, very readily, at the extremity of a Right Line; by a Scale of equal Parts. As BC, perpendicular to AB. e. g. Take, by any Scale, 3, 6, 9, or 12 equal Parts, which, fet off, from B to A; at which point, B, a Perpendicular to AB is required. Then take, for Radius, 4, [8, 12, or 16 Divifions, of the fame Scale; and, feting one Point of the Compaffes in B, make a small Ark, at C. By the fame Scale, take 5, 10, 15, or zo equal Parts, and with one point of the Compaffes, at A, cross the former Ark, at C, and draw BC; which will be perpendicular to AB. For, the Square of AC is equal to the two Squares of AB and BC, added together; by 20. 1. El. The Sides of two Squares being given, to find the Side of a Square which is equal to the difference between them; i. e. by how much the greater exceeds the lefs. X and Z are the Sides of the given Squares. Draw AC indefinite; in which, take AB equal X, and BC equal Z. Pr. 3. X E A Make ACD a R. Angle, and draw CD, indef. On B, with the Radius AB, defcribe the Ark AED, cuting CD at D; CD is the Side of a Square, equal to the difference between the Squares of AB and BC, or X & Z. Draw BD. DEM. BD (equal AB, equal X) is the Hypothenufe of the right-angled Triangle DCB; the fquare of which, is equal to the square of BC added to the fquare of CD. Confequently, BD fquare (equal X) exceeds BC fquare (equal Z) by the fquare of CD; by Prop. 20. 1. PRO B PROBLÉM XXX. 13. VI. Euclid. To find a mean Proportional between two given X and Z are the two given Lines. E Ꭰ Ᏼ " It is required to find a Line, to which, either of the Lines, X or Z, fhall have the fame Ratio or Proportion, as that Line has to the other. Or, the Square of which, fhall be equal to a Rectangle under the two given C Lines. Draw at pleasure A'C; make AB equal to one of the given DEM. For, draw AE and EC; AEC is a Right Angle--12.1.. SCHOL. This is the very fame, in the operation, as the 25th; for the PRO |