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PROP. XIII. PROB. Given the three perpendiculars of a triangle ; to construct it.
Let A, B, C be three given straight lines; it is required to describe a triangle having its three perpendiculars respectively equal to A, B, Č.
Take any straight line D, and describe a triangle EFG, having the sides FG, FE, EG third proportionals to A and D, B and D C and D; and draw the perpendiculars EH, GL, FK.
Then the rectangles FG.A, EF.B are equal, each being equal to the square of D; and therefore EF:FG::A: B. But in the similar triangles EHF, GLF, EF: FG :: EH: GL; wherefore EH : GL::A:B: and in the same manner it would be proved, that EH : FK :: A: C. Hence (V. 14.) if EH be equal to A, GL is equal to B, and FK to C; and EFG is the triangle required.
But if EH be not equal to A, make EM equal to it, and draw NMO parallel to FG, and meeting EF, EG, produced, if necessary, in N and 0 : ENO is the required triangle. Draw OP perpendicular to EN. Then EM : EH : : EO: EG, and OP: GL::EO : EG; whence (V. 11. and alternately,) EM: OP :: EH: GL; or, by the foregoing part, EM:0P :: A:B: wherefore, (V. 14.) since EM is equal to A, OP is equal to B : and it would be proved in a similar manner, that the perpendicular from N to EO is equal to C.
Method of Computation. By dividing any assumed number successively by A, B, C, we find the sides of the triangle EFG, and thence its angles, or those of ENO; whence, since the perpendicular EM is given, the sides are easily found. *
PROP. XIV. PROB.
Given the sum of the legs of a right-angled triangle, and the sum of the hypotenusey and the perpendicular to it from the right angle ; to construct the triangle.
Let the sum of the legs of a right-angled triangle be equal to
* Or, when the sides of EFG are found, its perpendicular EH may be computed in the manner pointed out in the note to the 13th proposition of the second book. Then EH: A:: FG: NO : : EF: EN:: EG: EO.
the straight line A, and the sum of the hypotenuse and perpendicular equal to BC: it is required to construct the triangle.
Find (I. 47. cor. 3.) a straight line the square of which is equal to the excess of the square of BC above that of A, and cut off BD equal to that line ; on DC as diameter describe a semicircle, and draw EF parallel to DC at a distance equal to BD; join either point of intersection, E, with D and C: B DEC is the required triangle.
Draw the perpendicular EG, which (const.) is equal to BD. Then (II. 4.) BC or A' + EG' = (DC + EG)' = DC + EG + 2DC.EG; whence A = DC + 2DC.EG. Also (DE + EC) = DE + EC' + 2DE.EC = DC + 2DC.EG, because (ITÍ. 31. and I. 47.) DC = DE + ECP, and DC.GE = DE.EC, each being equal to twice the area of the triangle DEC. Hence (DE + EC) = AR; wherefore (I. 46. cor. 3.) DE + EC = A; and DEC is the triangle required.
Method of Computation. From the construction, we have BD or EG = V(BC – A). Then DC = BC – BD; by halving which we get the radius of the semicircle : and if from the square of the radius drawn from E, the square of EG be taken, and if the square root of the remainder be successively taken from the radius, and added to it, the results will be the segments DG, GC; from which, and from EG, the sides (1. 47. cor. 1.) are readily computed.
PROP. XV. PROB. + Given the base of a triangle, the perpendicular, and the difference of the sides ; to construct it.
Make AB equal to the given base, and parallel to it draw CD, at a distance equal to the given perpendicular; draw BDF perpendicular to CD, and make DF equal to DB; from A as centre, with a radius AE equal to the given difference, describe the circle ELN, through B, F describe any circle cutting ELN in L, N, and let G be the point in which a straight line drawn through L, N cuts FB produced ; draw the tangent GK, and draw AKM cutting CD in M; join BM ; and it is evident, from the first corollary to the second proposition of the second book of this Appendix, that AMB is the required triangle.
Method of Computation. From M as centre, with MK as radias describe a circle, and by the corollary referred to, it will pass
through B and F. Join AG, and produce BA to H. Then the rectangle FG.GB = AGʻ— AK, each being equal to the square of GK; that is, FB.BG + BGʻ = AB + BG? - AK?. Take away BG? ; then FB.BG= AB — AK* = (AB+ AK).(ABAK) = HB.EB. Hence BG becomes known. Then, in the two right-angled triangles GBA, GKA, the angles at A can be computed, and their difference is the angle MAB in the required triangle. The rest is easy, if the perpendicular from M to AB be drawn.
PROP. XVI. PROB. GIVEN the base, the area, and the ratio of the sides of a triangle; to construct it.
Let AB be the given base, and (VI. 10. and schol.) find the points C, D, so that AC, CB and AD, DB are in the ratio of the sides ; on CD as diameter describe the semicircle CED, and (1. 45. cor.) to AB apply a 'parallelogram BÉ double of the given area; let FG, the side of this opposite to AB, produced if necessary, cut the semicircle in E, and join EA, EB: EAB is the required triangle.
For (VI. G. cor.) AE is to EB, as AC to CB ; that is in the given ratio. Also (1.41.) A EB is half of the parallelogram BF, which is double of the given area. Therefore AEB is on the given base, is of the given area, and has its sides in the given ratio. *
Schol. If FG produced do not meet the semicircle, the problem is impossible : if it cut it in E and E', there will be two triangles essentially different, each of which will answer the conditions of the problem : if it touch the semicircle, there will be only one triangle, and it will be the greatest possible with the base of the given magnitude, and the sides in the given ratio ; and hence we have the means of solving the problem in which it is required to construct a triungle on a given base, having its sides in a given ratio, and its area a maximum.
Method of Computation. Join the centre H with E, and draw the perpendicular EK. Then, let m:n :: AC : CB, and consequently m:n :: AD: DB; then, from these, by composition and division, we get m +n:n:: AB: CB,
m -n:n:: AB: DB; whence DC and its half, the radius of the circle, become known. EK also is found by dividing double the area by AB. Then, in the triangle EKH, KH can be found, and thence AK and KB; and, by means of them and EK, the sides EA, EB may be computed. If E' be taken as the vertex, the method of computation is almost the same, and is equally easy,
* See the Notes at the end of the volume.
PROP. XVII. PROB. Given the base of a triangle, the vertical angle, and the rectangle of the sides; to construct it.
On the given base describe a segment containing an angle equal to the given angle; to the diameter of the circle of which this segment is a part, and to the lines containing the given rectangle, find a fourth proportional : this proportional (VI. D.) is the perpendicular of the triangle ; and the rest of the solution is effected by means of the seventh proposition of this book.
PROP. XVIII. PROB. To divide a given triangle into two parts in a given ratio, by a straight line parallel to one of the sides.
Let ABC be a given triangle ; it is required to divide it into two parts in the ratio of the two straight lines m, n, by a straight line parallel to the side BC.
Divide (VI.10.) AB in G, so that BG:GA:: m:n, and between AB, AG find (VI. 13.) the mean proportional AH; draw HK parallel to BC: ABC is divided by HK in the manner required.
For (VI. 19. cor.) since the three straight lines AB, AH, AG are proportionals: AB is to AG, as the triangle ABC to AHK; whence, by division, BG is to AG, or (const.) m is to n, as the quadrilateral BCKH to the triangle AHK.
In practice, the construction is easily and elegantly effected, when the triangle is to be divided either into two or more parts proportional to given lines, by dividing AB into parts proportional to those lines, and through the points of section drawing perpendiculars to AB, cutting the arc of a semicircle described on AB as diameter : then by taking lines on AB, terminated at A, and severally equal to the chords drawn from A to the points of section of the arc, the points on AB will be obtained through which the parallels to BC are to be drawn. The reason is evident from the foregoing proof, in connexion with the corollary to the eighth proposition of the sixth book.
PROP. XIX. PROB. To divide a given triangle into two parts in a given ratio, by a straight line parallel to å given straight line.
Let ABC be a given triangle, and D a given straight line ; it
is required to divide ABC into parts in a given ratio, by a straight line parallel to D.
Divide AB in E, so that AE, EB may be in the given ratio, and draw EF parallel to D; take BG a mean proportional between BF and BC, and draw GH parallel to EF: GH divides ABC in the manner re. quired.
Join EC. Then (VI. 19. cor.) BF: BC:: BFE: BGH, and (VI. 1.) BF: BC::BFE: BCE; wherefore BFE : BGH :: BFE: BCE; and therefore BGH is equal to BEC; and consequently the remainders AHGC, AEC are equal. But (VI. 1.) BE: EA::BEC: AEC; therefore also BE: EA :: BGH: AHGC.
Should the point H be not in BA, but in its continuation, let CA, and not BA, be divided in the given ratio, and the rest of the construction will continue the same, the point C being used instead of B.
PROP. XX. PROB. To divide a triangle into two parts in a given ratio, by a straight line drawn through a given point in one of the sides.
Let ABC be the given triangle, and D the given point; and let AE be to EB in the given ratio ; join DC, draw EF parallel to it, and join DF: ABC is divided by DF in the given ratio.
Join EC. Then (1.37.) the triangles EFD, EFC are equal ; to each add BEF; then BDF, BEC are equal; and taking them separately from ABC, the remainders ADFC, A EC are equal. Now (VI. 1.) CEA, CEB are in the ratio of AE to EB, that is, in the given ratio ; and therefore ADFC, BDF are also in the given ratio.
PROP. XXI. PROB.
From a given point in one of the sides of a given triangle, to draw two straight lines trisecting the triangle.
Let ABC be the given triangle, and D the given point: then, if BD be not less than one-third of BC, to BD, to a third part of BC, and to the perpendicular from A to BC, find a fourth proportional, and at a distance equal to it draw a parallel to BC, cutting BA in E, and join ED: the triangle BED is evidently (VI. 16. and I. 41.) a third part of