Page images
PDF
EPUB

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.

E

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

D

G

[ocr errors]

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.

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.

D

K

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.

[ocr errors][merged small]

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.

[blocks in formation]

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.

[ocr errors]

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: GEM: 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

For, by construction DE.EF=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 AE=DE.GE; consequently,

QC AP: FG: GE:: M: N;

:

that is, the quadrilateral is divided, as required.

SUPPLEMENT

TO THE

ELEMENTS

OF

GEOMETRY.

21

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.

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

A

E

P

M

B

two points. The arcs of a circle are essentially convex ; but the present proposition extends to any line which fulfils the required conditions.

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.

« PreviousContinue »