F G line EC make (Prob. 9.) the angle CEF equal to D; and through A draw (Prob. 13.) AG parallel to BC, and through C draw CG (Prob. 13.) parallel to EF; therefore FECG is a parallelogram And because BE is equal to EC, the triangle ABE is likewise equal (32. 1.) to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC, AG; therefore the triangle ABC is double of the triangle AEC. And the parallelogram FECG is likewise double (35. 1.) of the triangle AEC, because it is upon the same base, and between the same parallels: Therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle D; Wherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of its angles CEF equal to the given angle D. B E D COR. Hence, if the angle D be a right angle, the parallelogram EFGC will be a rectangle, equivalent to the triangle ABC; and therefore the same construction will apply to the problem: to make a triangle equivalent to a given rectangle. PROP. XV. PROB. To a given straight line to apply a parallelogram, which shall be equivalent to a given triangle, and have one of its angles equal to a given rectilineal angle. 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. Make (Prob. 14.) the parallelogram BEFG equal to the triangle C, having the angle EBG equal to the angle D, and the side BE in the same straight line with AB: produce FG to H, and through A draw (Prob. 13.) AH parallel to BG or EF, and join HB. Then because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE, are together equal (21. 1.) to two right angles; wherefore the angles BHF, HFE are less than two right angles; But straight lines which with another straight line make the interior angles, upon the same side less than two right angles, do meet if produced (22. 1.): Therefore HB, FE will meet, if produced; let them meet in K, and through K draw KL parallel to EA or FH, and produce HA, GB to the points L, M: Then HLKF is a parallelogram, of which the diameter is HK, and AG, ME are the parallelograms about HK; and LB, BF are the complements; therefore LB is equal (36. 1.) to BF but BF is equal to the triangle C; wherefore LB is equal to the triangle C; and because the angle GBE is equal (8. 1.) to the angle ABM, and likewise to the angle D; the angle ABM is equal to the angle D: Therefore the parallelogram LB, which is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. COR. Hence, a triangle may be converted into an equivalent rectangle, having a side of a given length: for, if the angle D be a right angle, and AB the given side, the parallelogram ABML will be a rectangle equivalent to the triangle C. PROP. XVI. PROB. To describe a parallelogram equivalent to a given rectilineal figure, and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Join DB, and describe (Prob. 14.) the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH (Prob. 15.) apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. And because the angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM; add to each of these the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM; but FKH, KHG are equal (21. 1.) to two right angles; 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, upon the opposite sides of GH, make the adjacent angles equal to two right angles, KH is in the same straight line (7. 1.) with HM. And because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal (21. 1.); add to each of these the angle HGL; therefore the angles MHG, HGL, are equal to the angles HGF, HGL: But the angles MHG, HGL, are equal (21. 1.) to two right angles; wherefore also the angles 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 (23. 1.) to ML; but 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 rectilineal figure ABCD is equal to the whole parallelogram KFLM; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. 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 (Prob. 15.) to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle. PROP. XVII. PROB. To find a triangle that shall be equivalent to any given rectilineal figure. Let ABCDE be the given rectilineal figure. Draw the diagonal CE, cutting off the triangle CDE; draw DF parallel to CE, meeting AE produced, and join CF: the polygon ABCDE will be equivalent to the polygon ABCF, which has one side less than the original polygon. For the triangles CDE, CFE, have the base CE common, and they are between the same parallels; since their vertices D, F, are situated in a line DF parallel to the base: these triangles are therefore equivalent. Draw, now, the diagonal CA and BG parallel to it, meeting EA produced: join CG, and the polygon ABCF will be reduced to an equivalent triangle; and thus the pentagon ABCDE B C D will be reduced to an equivalent triangle GCF. The same process may be applied to every other polygon; for, by successively diminishing the number of its sides, one being retrenched at each step of the process, the equivalent triangle will at length be found. COR. Since a triangle may be converted into an equivalent rectangle, it follows that any polygon may be reduced to an equivalent rectangle. PROP. XVIII. 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. E From the point A draw (Prob. 6.) AC at right angles to AB; and make (Prob. 3.) AD equal to AB, and through the point D draw DE parallel (Prob. 13.) to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram; whence AB is equal (28. 1.) to DE, and AD to BE; but BA C is equal to AD: therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equila- D teral; it is likewise rectangular; for the straight line AD meeting the parallels AB, DE, makes the angles BAD, ADE equal (21. 1.) to two right angles; but BAD is a right angle; therefore also ADE is a right angle; now the opposite angles of parallelograms are equal (28. 1.); therefore each of the opposite angles ABE, BED is a right angle; where A B fore 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. PROP. XIX. PROB. To find the side of a square that shall be equivalent to the sum of two squares. Draw the two indefinite lines AB, AC, perpendicular to each other. Take AB equal to the side of one of the given squares, and AC equal to the other; join BC: this will be the side of the square required. For the triangle BAC being right angled, the square constructed upon BC (Th. 37. 1.) is equal to the sum of the squares described upon AB and AC. SCHOLIUM. B A square may be thus formed that shall be equivalent to the sum of any number of squares; for a similar construction which reduces two of them to one, will reduce three of them to two, and these two to one, and so of others. PROP. XX. PROB. To find the side of a square equivalent to the difference of two given squares. Draw, as in the last problem, (see the fig.) the lines AC, AD, at right angles to each other, making AC equal to the side of the less square; then, from C as centre, with a radius equal to the side of the other square, describe an arc cutting AD in D: the square described upon AD will be equivalent to the difference of the squares constructed upon AC and CD. For the triangle DAC is right angled; therefore, the square described upon DC is equivalent to the squares constructed upon AD and AC: hence (Cor. 1. Th. 37. 1.), AD2=CD2—AC2. PROP. XXI. PROB. A rectangle being given, to construct an equivalent one, having a side of a given length. Let AEFH be the given rectangle, and produce one of its sides, as AH, till HB be the given length, and draw BFD meeting the prolongation of AE in D; then produce EF till FG is equal to HB: draw BGC, HFK, parallel to AED, and through the point D draw DKC parallel to AB or EG; then, the rectangle GFKC, having the side FG of a given length, is equal to the given rectangle AEFH (Th. 36. 1.). A E H F B G COR. A polygon may be converted into an equivalent rectangle, having one of its sides of a given length. |