SOLUTION. Let ABC be the given triangle, which is required to be converted into a trapezoid, one of whose parallel sides is the base of the triangle; further, let the angle ABC of the triangle be also one of the angles of the trapezoid, and the other angle at C, equal to the given angle m.

1. On BC make the angle BCD equal to the angle m (problem Vl.)

2. From the vertex A of the triangle, draw the line AD parallel to BC, which meets CD in D, and from B draw the line BE parallel to CD, which meets AD in E.

3. Upon AD describe a semicircle AFD, and from E draw the perpendicular EF, meeting the semicircle in F.

4. From D, with the radius DF, describe the arc FG, meeting the line AD in G.

5. From G draw the line GH, parallel to DC, and from the point H, in which it meets AB, the line HI, parallel to BC; then BHIC is the trapezoid sought.

DEMON. Suppose AF drawn. The triangle AFD is rightangular in F (remark page 198); therefore the side DF is a mean proportional between the whole hypothenuse AD and the part ED (page 89, 2dly); that is, we have the proportion

AD: DFDF : DE,

or, since DF is equal to DG,

AD: DGDG: DE;

and as in every geometrical proportion the second term can be subtracted from the first term, if the fourth be also subtracted from the third term, without destroying the proportion, we have also AD-DG: DG = DG — DE: DE,

and by changing the order of the mean terms (principle 2d of geom. prop. page 67)

ADDG: DG-DE=DG: DE,

which may also be written

AG: GE = HI: BC;

(because AD less DG is AG; DG less DE is GE; DG is by construction equal to HI, and DE is equal to BC.)

Further, from the two similar triangles AHG, ABE, (the line HG is drawn parallel to BE), we have the proportion AB: AH = = AE: AG;

and consequently,

ABAH AH = AE · - AG: AG;

or, HB : AH = GE: AG;

(because ABAH is HB, and AE AG is GE).

And by changing the order of the

ciple 1st of geom. prop. page 67)

terms in each ratio (prin

AH: HBAG: GE.

This last proportion and the above one, AG: GE=HI : BC, have the ratio AG: GE common; consequently we have the new proportion

AH: HB: - HI: BC.

Suppose HC drawn. The two triangles AHC, HBC, have the same height MC, consequently their areas are to each other as the bases AH, HB, (see page 111, 6thly), and we have the proportion

triangle AHC triangle BCH : For the same reason we have

triangle HIC triangle BCH :

AH: HB.

= HI: BC;

(because the two triangles HIC, BCH, being, by construction 5, between the same parallels, have the same height CO); and as we have the proportion

we have also

AH: HB

HI: BC,

triangle AHC: triangle BCH = triangle HIC : triangle BCH. This last proportion expresses, that the area of the triangle AHC is as many times greater than the area of the triangle BCH, as the area of the triangle HIC is greater than the area of the same triangle BCH; consequently the triangle AHC is in area equal to the triangle HIC; and therefore, adding to each of the two equal triangles AHC, HIC, the same triangle HBC, the area of the whole triangle ABC is equal to that of the trapezoid HIBC.

PROBLEM XXXIV. To convert a given trapezoid into another, having one side, and one of the angles adjacent to this side, common with the former, and the other angle equal to a given

SOLUTION. Let ABCD be the trapezoid to be converted into another trapezoid, which has the

angle at A and the side AD common; but the other angle at D equal to the given angle m.

Convert the trapezoid ABCD into the triangle AED, having the same base AD and the angle at A common (see the rule given in problem XII); then this triangle again into a trapezoid ADBC, so that the angle ADG is equal to the given angle m; the trapezoid ADBC is the one sought.

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PART III.

Partition of figures by drawing.

PROBLEM XXXVI. To divide a triangle from one of the vertices into a given number of parts.

SOLUTION. Let ABC be the given triangle, which is to be divided, say, into six equal parts; let A be the vertex, from which the lines of division are to be drawn.

1. Divide the side

BC, opposite the vertex A, into six equal parts, BD, DE, EF, FG, GH, HC.

2. From A to the

A

B D E F G H C

points of division D, E, F, G, H, draw the lines AD, AE, AF, AG, AH; the triangle ABC is

divided into the six equal triangles, ABD, ADE, AEF, AFG, AGH, AHC.

DEMON. The triangles ABD, ADE, AEF, AFG, AGH, AHC, are, in area, equal to one another, because they have equal bases and the same height Am (page 108).

Remark. If it is required to divide the triangle ABC according to a given proportion, it will only be necessary to divide the line BC in this proportion, and from A to draw lines to these points of division.

PROBLEM XXXVII. From a given point in one of the sides of a triangle, to divide it into a given number of equal parts.

SOLUTION. Let ABC be the given triangle, which is to be divided into eight equal parts; the lines of division are to be drawn from T.

a

A

E

F

G

B K

I

H

1. Make Aa and Bb equal to of AB, and from T draw the line TC to the vertex C of the triangle.

2. From a and b draw the lines a D, b K parallel to TC, meeting the sides AC, BC, in D and K.

3. From AC, from A towards C, measure off the distance AD as many times as possible (in this case four times); and thus determine the points E, F, G; from BC, in the direction from B towards C, also measure off the distance BK