rhomboid AM is divided into equal triangles, as also the rhomboids DL, EK, BI, and FC, and each of these rhomboids contain one or more of the triangles contained in DK; hence the triangles are all equal, and they are similar to ABC.

PROPOSITION XXVI. PROBLEM.

To divide a quadrilateral into two parts by a straight line drawn from C, 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 A P equal to twice EG, and join

BE G

PC, then the quadrilateral will be divided as required.

F

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 to the quadrilateral DP, and GF is to EG as M is to Ñ, therefore DP CPB :: M : N;

that is, the quadrilateral is divided as required.

PROPOSITION XXVII. PROBLEM.

To divide a quadrilateral AC into two parts by a line parallel to AB one of its sides, so that these parts may be to each other as the line M is to the line N.

as AB is to KB, so make DA to Eas; draw ab parallel to AB, and it will divide the quadrilateral into the required parts.

For since the triangles EAB, Eab are similar, we have the proportion EAB: Eab: EA: Ea; 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 by construction, AK: KH:: M: N; consequently, Ab aC : M: N;

D

that is, the quadrilateral is divided as required.
If the sides AD, BC are parallel,
then make AE=BC, bisect ED in F,
and divide AF in a, so that Aa: aF
:: M: N, then the parallel ab will
divide the quadrilateral as required.

For draw FG parallel to ab meeting the production of BC, then since FD=CG, the triangles DFH, GCH are equal; so that the quadrilateral aC is equivalent to therhomboid aG, and by construction M: N:: Aa: aF:: Ab: aG; consequently,

MN: Ab : aC.

a

E

F

To divide a quadrilateral into two parts by a line drawn from P, a point in one of its sides, so that the parts be to each other as a line M is to a line N.

may

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 LK=PBCp; but PD-DL: PD·LK :: DL: LK :: M: N; consequently,

APPD PBCp :: M: N;

hence the quadrilateral is divided as required.

PROPOSITION XXIX. PROBLEM.

To divide a quadrilateral ABCD by a line perpendicular to one of the sides AB, so that the two parts may be to each other as a line M is to a line N.

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 (Prop. XXVI.) divide the quadrilateral EC into F two parts by a line PQ, pa

GA

D

PC

E

Q

B

rallel to the side 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.

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 AP=DE GE; consequently,

QC AP:: FG: GE :: M: N; that is, the quadrilateral is divided as required.

PROPOSITION XXX. PROBLEM.

To divide a quadrilateral ABCD into three equivalent parts by lines drawn from C, the vertex of one of its angles.

Draw the diagonal AC, and the perpendicular CE, which bisect in F; construct upon FE a rectangle equivalent to the triangle DAC, and let AG be equal to the other side of this rectangle.

Take AP equal to a third part of AB-2AG; bisect PB in P'; then draw the lines CP, CP', and they will divide the quadrilateral into three equivalent parts.

=

For PB-AB-AP, and since AP=} AB- AG, by construction, it follows that PB AB — (÷AB-AG)=3AB+ AG, that is to say, PB 2 GB, and therefore PB-FE GB FE, that is, the triangle CPB is two thirds of the quadrilateral AC; but the triangles CPP', CP'B having equal bases, PP', P'B, are each half the triangle CPB, and consequently one third of the quadrilateral: hence the spaces DAPC, CPP', CP'B are equivalent.

2

Scholium.

In this case of the problem the lines of division must necessa rily fall within the triangle CAB; for AB being greater than 2AG, AB·FE>2AG:FE, that is, the triangle CAB exceeds twice the triangle DAC, and is therefore greater than two thirds of the quadrilateral. In the third case of the problem (pro vided AG is not equal to AB, nor to 2AB) this triangle will be less than two thirds of the quadrilateral, but greater than one third; and consequently one line of division only will fall within the triangle CAB, and the other within the triangle DAC; in this case, therefore, after having determined the line CP', by making AP' (2AB-AG), it will only be necessary to divide the remaining quadrilateral AP'CD, by proposition XXVI., into two equal parts, by a line from the point C. In the second case of this problem, that is to say, when AB is less than half AG, the triangle CAB will be less than one third of the proposed quadrilateral, and consequently both lines of division will fall within the other triangle DAC; and therefore this case is virtually the same as the first; the perpendicular from C being to the side AD instead of AB. When we happen to have AG AB, or AG=2AB, then the diagonal CA will be one line of division.

A mere inspection of the figure will always enable us to determine upon which side the perpendicular is to fall; for as one triangle will be always at least double the other in surface, the greatest will be at once recognized:

PROPOSITION XXXI. PROBLEM.

To divide a quadrilateral AD into three equivalent parts by lines drawn from a given point P. in one of its sides.

the altitude of the former be less than half the altitude of the latter, the triangle CPB will obviously be less than one third of the quadrilateral AC; and if it be greater, then the same triangle will exceed one third of the quadrilateral. Suppose then CPBAC. Draw PE perpendicular to AB, and upon FE, the half of PE, construct a rectangle equivalent to ADPB, and let the other side of this rectangle be equal to GB; construct also on FE a rectangle equivalent to the triangle CPB, and let BH be the other side of this rectangle.

Take BKGB-3 BH, join PK, then PK will be one of the lines of division.

For the triangle PKB is equal to the rectangle KB·FE, and the triangle CPB=BH FE by construction, therefore, PK BCP=(KB+BH) FE; but by construction, KB= } GB— BH; therefore KB + BHGB + } BH = } GH; hence PKBCPGH FE ABCD.

-

If CPB > AC, then, instead of the perpendicular PE to AB, draw PE perpendicular to BC or its production, and upon F' E', the half thereof, construct a rectangle equivalent to ADPB, and let G'B be equal to its base. Take BK BCBG', and join PK', then PK will be one of the lines of di

vision.