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For, because AB : CD:: CE: AG, by Prop. I., B. II., ABX AG=CD X CE. Therefore the rectangle ABHG is equivalent to the rectangle CDFF; and it is constructed upon the given line AB.
To construct a triangle which shall be equivalent to a given polygon. Let ABCDE be the given polygon; it
D is required to construct a triangle equiva'ent to it.
Draw the diagonal BD cutting off the E triangle BCD. Through the point C, draw CF parallel to DB, meeting AB
А produced in F. Join DF; and the polygon AFDE will be equivalent to the polygon ABCDE.
For the triangles BFD, BCD, being upon the same base BD, and between the same parallels BD, FC, are equivalent. To each of these equals, add the polygon ABDE; then will the polygon AFDE be equivalent to the polygon ABCDE; that is, we have found a polygon equivalent to the given polygon, and having the number of its sides diminished by
In the same manner, a polygon may be found equivalent to AFDE, and having the number of its sides diminished by one; and, by continuing the process, the number of sides may be at last reduced to three, and a triangle be thus obtained equivalent to the given polygon.
To make a square equivalent to the sum or difference of two given squares.
First. To make a square equivalent to the sum given squares. Draw two indefinite lines AB, BC at right angles to each other. Take AB equal to the side of one of the given squares, and BC equal to the side of the other. Join AC; it will be the side of the A required square.
For the triangle ABC, being right-angled at B, the square
on AC will be equivalent to the sum of the squares upon AB and BC (Prop. XI., B. IV.).
Secondly. To make a square equivalent to the difference of two given squares. Draw the lines AB, BC at right an gles to each other; and take AB equal to the side of the less square. Then from A as a center, with a radius equal to the side of the other square, describe an arc intersecting BC in C; BC will be the side of the square required; because the square of BC is equivalent to the difference of the squares of AC and AB (Prop. XI., Cor. 1, B. IV.).
Scholium. In the same manner, a square may be made equivalent to the sum of three or more given squares; for the same construction which reduces two of them to one will reduce three of them to two, and these two to one.
Upon a given straight line, to construct a polygon simila to a given polygon. Lou ABCDE be the giv
D en polygon, and FG be the given straight line; it E is required upon the line FG to construct a polygon similar to ABCDE. Draw the diagonals BD,
G At the point F, in the straight line FG, make the angle GFK equal to the angle BAE; and at the point G make the angle FGK equal to the angle ABE. The lines FK, GK will intersect in K, and FGK will be a triangle similar to ABE. In the same manner, on GK construct the triangle GKI similar to BED, and on GI construct the triangle GIH similar to BDC. The polygon FGHIK will be the polygon required. For these two polygons are composed of the same number of triangles, which are similar to each other, and similarly situated; therefore the polygons are similar (Prop. XXV., Cor., B. IV.)
Given the area of a rectangle, and the sum of two adjacent sides, to construct the rectangle.
Let AB be a straight line equal to the sum of the sides of the required rectangle.
Upon AB as a diameter, describe a c semicircle. At the point A erect the perpendicular AC, and make it equal to the side of a square having the given area. Through C draw the line CD
E B allel to AB, and let it meet the circumference in D; and from D draw DE perpendicular to AB. 1 hen will AE and EB be the sides of the rectangle required.
For, by Prop. XXII., Cor., B. IV., the rectangle AE XEB is equivalent to the square of DE or CA, which is, by construction, equivalent to the given area. Also, the sum of the sides AE and EB is equal to the given line AB.
Scholium. The side of the square having the given area, must not be greater than the half of AB; for in that case the line CD would not meet the circumference ADB.
Given the area of a rectangle, and the difference of two adjacent sides, to construct the rectangle.
Let AB be a straight line equal to the difference of the sides of the required rectangle.
Upon AB as a diameter, describe a circle; and at the extremity of the diameter, draw the tangent AC equal to the side of a square having the given area. Through the point C and the center F draw the secant CE; then will CD, CE be the adjacent sides of the rectangle required.
For, by Prop. XXVIII., B. IV., the rectangle CD XCE is equivalent to the square of AC, which is, by construction, equivalent to the given area. Also, the difference of the lines CE, CD is equal to DE or AB.
REGULAR POLYGONS, AND THE AREA OF THE CIRCLE.
A regular polygon is one which is both equiangular anu equilateral.
An equilateral triangle is a regular polygon of three sides; a square is one of four.
Regular polygons of the same number of sides are similar figures.
Let ABCDEF, abcdef be
E two regular polygons of the same number of sides; then will they be similar figures. A
d For, since the two polygons have the same number of
B sides, they must have the same number of angles. Moreover, the sum of the angles of the one polygon is equal to the sum of the angles of the other (Prop. XXVIII., B. I.); and since the polygons are each equiangular, it follows that the angle A is the same part of the sum of the angles A, B, C, D, E, F, that the angle a is of the sum of the angles a, b, c, d, e, f. Therefore the two angles A and a are equal to each other. The same is true of the angles B and b, C and c, &c.
Moreover, since the polygons are regular, the sides AB, BC, CD, &c., are equal to each other (Def.); so, also, are the sides ab, bc, cd, &c. Therefore AB: ab :: BC:bc:: CD:cd, &c. Hence the two polygons have their angles equal, and their homologous sides proportional; they are consequently similar (Def. 3, B. IV.). Therefore, regular polygons, &c.
Cor. The perimeters of two regular polygons of the same Humber of sides, are to each other as their homologous sides, and their areas are us the squares of those sides (Prop. XXVI., B. IV.).
Scholium. The angles of a regula: polygon are determined by the number of its sides.
A circle may be described about any regular polygon, an another
may be inscribed within it. Let ABCDEF be any regular polygon; a circle may be described about it, and another may be inscribed within it.
Bisect the angles FAB, ABC by the A straight lines AO, BO; and from the point
Then, because in the triangles OBA, OBC, AB is, by hypothesis, equal to BC, BO is common to the two triangles, and the included angles OBA, OBC are, by construction, equal to each other; therefore the angle OAB is equal to the angle OCB. But OAB is, by construction, the half of FAB; and FAB is, by hypothesis, equal to DCB; therefore OCB is the half of DCB, that is, the angle BCD is bisected by the line OC. In the same manner it may be proved that the an gles CDE, DEF, EFA are bisected by the straight lines OD, OE, OF.
Now because the angles OAB, OBA, being talves of equal angles, are equal to each other, OA is equal to OB (Prop. XI., B. I.). For the same reason, OC, OD, OE, OF are each of them equal to OA. Therefore a circumference described from the center 0, with a radius equal to OA, will pass through each of the points B, C, D, E, F, and be described about the polygon.
Secondly. A circle may be inscribed within the polygon ABCDEF. For the sides AB, BC, CD, &c., are equa chords of the same_circle; hence they are equally distant from the center O (Prop. VIII., B. III.) ; that is, the perpendiculars OG, OH, &c., are all equal to each other. Therefore, if from O as a center, with a radius OG, a circumference be described, it will touch the side BC (Prop. IX., B. III.), and each of the other sides of the polygon; hence the circle will be inscribed within the polygon. Therefore a circle may be described, &c. Scholium 1. In regular polygons, the center of the inscribed