2. Bisect the angle DAE by the line AM (Problem III.); and the right angle BAC is divided into the three equal angles CAE, EAM, MAB.

DEMON. The angle BAE being one of the angles of an equilateral triangle, is one third of two right angles (pages 44 and 36), and therefore two thirds of one right angle; consequently CAE is one third of the right angle BAC; and since the angle BAE is bisected by the line AM, the angles EAM, MAB, are each of them also equal to one third of a right angle; and are therefore equal to the angle CAE and to each other.

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PART II.

Problems relative to the transformations of geometrical figures.

PROBLEM XXIII. To transform a given quadrilateral figure into a triangle of equal area, whose vertex shall be in a given angle of the figure, and whose base in one of the sides of the figure.

SOLUTION. Let ABCD (fig. I. and II.) be the given quadrilateral; the figure I. has all its an

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gles out wards, and the figure II. has one angle BCD in wards; let the vertex of the triangle, which shall be equal to it, fall in B.

1. Draw the diagonal BD (fig. I. and II.) and from C, parallel to it, the line CE.

2. From E, where the line CE cuts AD (fig. II.), or its further extension (fig. I.), draw the line EB; the triangle ABE is equal to the quadrilateral ABCD.

DEMON. The area of the triangle BCD (fig. I. and II.) is equal to the area of, the triangle BDE; because these two triangles are upon the same basis BD, and between the same parallels BD, CE (page 110, 3dly); consequently (fig. I.) the sum of the areas of the two triangles ABD and BDC is equal to the sum of the areas of the two triangles ABD, BDE; that is, the area of the quadrilateral ABCD is equal to the sum of the areas of the two triangles AF¿D, BDE, which is the area of the triangle ABC.

And in figure, II. the difference between the areas of the two triangles ABD, BCD; that is, the quadrilateral ABCD is equal to, the difference between the triangles ABD, EBD, which is the triangle ABE.

PROBLEM XXIV. To transform a given pentagon into a triangle, whose vertex shall be in a given angle of the pentagon, and whose base upon one of its sides.

SOLUTION. Let ABCDE (fig. I. and II.) be the given pentagon; let the vertex of the triangle, which is to be equal to it, be in C.

1. From C draw the diagonals CA, CE.

2. From B draw BF parallel to CA, and from D draw DG parallel to CE.

3. From F and G, where these parallels cut AE or its further extension, draw the lines CF, CG, CFG is the triangle sought.

DEMON. In both figures we have the area of the triangle CBA equal to the area of the triangle CFA; because these two triangles are upon the same basis CA, and between the same parallels AC, FB; and for the same reason is the area of the triangle CDE equal to the area of the triangle CGE; therefore in figure I. the sum of the areas of the three triangles CAE, CBA, CDE, is equal to the sum of the areas of the triangles CAE, CFA, CGE; that is, the area of the pentagon ABCDE is equal to the area of the triangle CFG: and in figure II. the difference between the area of the triangle CAE and the sum of the areas of the two triangles CBA, CDE, is equal to the difference between the area of the same triangle CAE and the sum of the areas of the two triangles CFA, CGE; that is, the area of the pentagon ABCDE is equal to the area of the triangle CFG.

PROBLEM XXV. To convert any given figure into a triangle, whose vertex shall be in a given

angle of the figure, and whose base is one of

Let ABCDEF (fig. I. and II.) be the given figure (in this case a hexagon,) and A the angle in which the vertex of the required triangle shall be situated. For the sake of perspicuity I shall enumerate the angles and sides of the figure from A, and call the first angle A, the second B, the third C, and so on; further, AB the first side, BC the second, DE the third, and so on. We shall then have the following general solution.

1. From A to all the angles of the figure draw the diagonals AC, AD, AE, which, açcording to the order in which they stand here, call the first, second, and third diagonal.

2. Draw from the second angle B, a line B a parallel to the first diagonal AC; from the point a, where the parallel meets the third side CD, (fig. II.), or its further extension (fig. I.), draw a line a b parallel to the second diagonal AD; and from the point b, where this meets

the fourth side DE (fig. II.) or its further extension (fig. I.), draw another line b c parallel to the third diagonal.

3. When in this way you have drawn a parallel to every diagonal, then from the last point of section of the parallels and sides, (in this case c), draw the line cA; A c F is the required triangle, whose vertex is in A, and whose base is in the side EF.

The demonstration is similar to the one given in the two last problems (pages 203, 204.) First, each of the hexagons is converted into the pentagon A a DEF, then the pentagon A a DEF into a quadrilateral Ab EF, and finally the quadrilateral into the triangle Ac F. The areas of these figures are evidently equal to one another; for the areas of the triangles, which, by the above construction, are successively cut off from these figures, are equal to the areas of the new triangles, which are successively added to them. (See the demonstration of the last problem.)

Remark. Although the solution given here, is only intended for a hexagon, yet it may easily be applied to every other rectilinear figure. All depends upon the substitution of one triangle for another by means of parallel lines; in which you have only to take care, that one side of the triangle substituted, be in a side of the figure, or in its further extension; because by these means the number of sides will be diminished. Moreover it is not absolutely necessary actually to draw the parallels; it is only requisite to note the points in which they cut the sides or their further extension; because all depends upon the determination of these points of section.

PROBLEM XXVI. To transform any given figure into a triangle whose vertex shall be in a certain point, in one of the sides of the figure, or