PROP. U. PROB. To divide a quadrilateral into two parts by a straight line drawn from the vertex of one of its angles, so that the parts may be to each other as a line M to another line N. D Draw CE perpendicular to AB, and construct a rectangle equivalent to the given quadrilateral, of which one side may be CE; let the other side be EF; and divide EF in G, so that M:N::GF: EG; take BP equal to twice EG, and join PC, then the quadrilateral will be divided as required. For, by construction, the triangle CPB is equivalent to the rectangle CE.EG; therefore the rectangle CE, GF is to the triangle CPB as GF is to EG. Now CE.GF is equivalent P BEG F to the quadrilateral DP, and GF is to EG as M is to N; therefore, DP: CPB:: M: N; that is, the quadrilateral is divided, as required. PROP. W. PROB. To divide a quadrilateral into two parts by a line parallel to one of its sides so that these parts may be to each other as the line M is to the line N. Produce AD, BC till they meet in E; draw the perpendicular EF and bisect it in G. Upon the side GF construct a rectangle equivalent to the triangle EDC, and let HB be equal to the other side of this rectangle. Divide AH in K, so that AK : KH :: M: N, and as AB is to KB, so make EA2 to Ea2; draw ab parallel to AB, and it will divide the quadrilateral into the required parts. A E C D G b K F H B For since the triangles EAB, Eab are similar, we have the proportion EAB Eab: EA2: Ea2; but by construction, EA2: Ea2 :: AB : KB; so that EAB: Eab :: AB: KB:: AB.GF: KB.GF; and consequently, since by construction EAB=AB.GF, it follows that Eab=KB. GF, and therefore AK.GF=Ab, and since by construction AH.GF=AC, it follows that KH.GF=aC. Now AK.GF: KH.GF:: AK: KH; but AK: KH:: M: N; consequently, Ab: aC M: N; that is, the quadrilateral is divided, as required. PROP. X. PROB. F K To divide a quadrilateral into two parts by a line drawn from a point in one of its sides, so that the parts may be to each other as a line M is to a line N. Draw PD, upon which construct a rectangle equivalent to the given quadrilateral, and let DK be the other side of this rectangle; divide DK in L, so that DL: LK :: M: N; make DF=2DL, and FG equal to the perpendicular Aa; draw Gp parallel to DP; join the points P, p, and the quadrilateral figure will be divided, as required. = For draw the perpendicular pb; then by construction, PD.DK = AC, and PD.DF PD.Aa + PD.pb, that is, PD.DF is equivalent to twice the sum of the triangles APD, pPD, consequently, since DL is half DF, PD.DL=APpD; and therefore PD. D LK-PBCp; but PD.DL: PD.LK:: DL: LK:: M: N; consequently, APPD: PBCp:: M: N; hence the quadrilateral is divided, as required. PROP. Y. PROB. To divide a quadrilateral by a line perpendicular to one of its sides, so that the two parts may be to each other as a line M is to a line N. Let ABCD be the given quadrilateral, which is to be divided in the ratio of M to N by a perpendicular to the side AB. Construct on DE perpendicular to AB, a rectangle DE.EF, equivalent to the quadrilateral AC, and divide FE in G, so that FG: GE:: M: N. Bisect AE in H, and divide the quadrilateral EC into two parts by a line PQ, parallel to DE, so that those parts may be to each other as FG is to GH, then PQ will also divide the quadrilateral AC as required. D Po FGA HE Q B For, by construction DE.LF=AC, and DE.EH-DAE; hence DE. HF=EC, and consequently, since the quadrilateral EC is divided in the same proportion as the base FH of its equivalent rectangle, it follows that QC=DE.FG, and EP-DE.GH, also AP=DE.GE; consequently, QC: AP:: FG: GE:: M: N; that is, the quadrilateral is divided, as required. ELEMENTS OF GEOMETRY. SUPPLEMENT. BOOK I. OF THE QUADRATURE OF THE CIRCLE. LEMMA Any curve line, or any polygonal line, which envelopes a convex line from one end to the other, is longer than the enveloped line. Let AMB be the enveloped line; then will it be less than the line APDB which envelopes it. M We have already said that by the term convex line we understand a line, polygonal, or curve, or partly curve and partly polygonal, such that a straight line cannot cut it in more than two points. If in the line AMB there were any sinuosities or re-entrant portions, it would cease to be convex, because a straight line might cut it in more than two points. The arcs of a circle are essentially convex; but the present proposition extends to any line which fulfils the required conditions. A B This being premised, if the line AMB is not shorter than any of those which envelope it, there will be found among the latter, a line shorter than all the rest, which is shorter than AMB, or, at most, equal to it. Let ACDEB be this enveloping line: any where between those two lines, draw the straight line PQ, not meeting, or at least only touching, the line AMB. The straight line PQ is shorter than PCDEQ; hence, if instead of the part PCDEQ, we substitute the straight line PQ, the enveloping line APQB will be shorter than APDQB. But, by hypothesis, this latter was shorter than any other; hence that hypothesis was false; hence all of the enveloping lines are longer than AMB |