From the extremity B, of the base AB, Fig. 68, draw B C at right angles to A B; set off B C equal to A B, and C D equal to CA; complete the rectangle ABDE, and draw its diagonals intersecting in F; with centre F describe a circle about the rectangle; from the angular points of the rectangle set off the distance A B to the points G H K, and L, on the circumference; and, joining the adjacent points, the required octagon will be described.

Fig. 68.

Reason. Since A C is equal to CD, the angles CAD and CDA are

equal (Theor. 4), and the angle ACB is

therefore double the angle CD A (Theor. 5). In like manner the angle A F B is also double the angle CD A, and therefore equal to the angle A C B.

Again, ABC being a right angle, B A C and ACB are together equal to a right angle (Theor. 5); and they are also equal to one another; therefore, each of them is half a right angle; and hence the angle A F B is also half a right angle. But the triangles B F G and D F H are each made equal in all respects to the triangle AF B; therefore the angles B F G, and D F H, are also each half a right angle, and therefore the remaining angle H F G is also half a right angle. In like manner the other four angles at F are each half a right angle; hence the eight triangles, having their vertices at F, have their vertical angles, and the sides containing these angles, respectively, equal to each other, and are therefore equal in all respects, and consequently the sides of the octagon are all equal to each other, and so also are its angles. To check the accuracy of the construction, see that L G and KH are each equal to A E or B D.

99. PROBLEM 30.-On a given base to construct an equilateral and equiangular polygon, or many-sided figure, whatever be the number of sides.

Divide 360° by twice the number of sides; subtract the quotient from 90°, and the remainder will be the measure of each of the angles between the sides of the figure and the radii of the circumscribing circle. Hence setting off this measurement from either end of the base, the lines thus drawn will intersect at the centre of the circumscribing circle.

For example, let it be required to describe a regular heptagon, or seven-sided figure, on a base two-thirds of an inch in length.

360°

14

25 = 64.

Set off, therefore, at A and B, the angles BA C, and A B C, each equal to 64; then C, the intersection of A C and B C, is the centre of the circle circumscribing the required heptagon; with centre C and radius CA or CB, describe a circle, step round it with the distance A B, and the required heptagon will be described.

is equal to

; hence

7

there are seven such angles

contained in a complete revolution.

Fig. 69.

From the difficulty of setting off the angle 64% with sufficient exactness, the points of the compasses, most probably, will not at first come round to the points from which they started, and the position of C will have to be slightly altered; but after two or three trials the object will be accomplished.

100. PROBLEM 31.-To construct a trapezoid of given dimen

sions.

Let A B C D, Fig. 70, be any trapezoid; draw B E parallel to D C; then the problem is reduced to the construction of the triangle A B E, and the completion of the parallelogram EBCD.

Let the lengths of the parallel sides be 5 feet and 2 feet, and the lengths of the slopes 4 feet and 3 feet.

Draw the straight line A D, Fig, 71, 5 feet long: set off 2 feet from D to E; from centres A and E, with radii of 4 feet

and 3 feet, respectively, strike arcs intersecting in B; join A B, B E, and complete the parallelogram E B C D.

Let the lengths of the parallel sides be 7 feet and 3 feet, and the inclinations of the slopes be 60° and 45°.

Draw A D, Fig. 72, 7 feet long, and set off D E equal to 3 feet; make the angles E A B, and A E B, equal to 60° and 45°, respectively, and complete the parallelogram E B C D.

Let the width of a ditch be 8 feet, its depth 3 feet, and the inclination of the slopes 42° and 54°.

Draw A D, Fig. 73, and FB at right angles to one another; set off F B equal to 3 feet; make the angles F BA, and F BE

equal to 36° and 48°, the complements of 54° and 42°, respectively; set off A D equal to 8 feet, and complete the parallelogram EBCD.

101. Every rectilineal figure can be divided into triangles, and in order to construct it sufficient data must be given for the construction of each of these triangles.

102. EXERCISES ON CHAPTER II.

(1.) A man walks due east 3 miles, and then south 2 miles. Construct a figure and find how far he is from home.-Ans. 3.6 miles.

(2.) Construct a square on a line 5 inches long; take a point on one of its diagonals at 3 inches distance from one of the angles through which it is drawn, and find the distances of this point from each of the other angles of the square.—Ans. 4.07, 3.58, and 3.58 inches.

(3.) Draw a straight line A B, whose length is limited by two straight lines drawn through a point P, 2.2 inches above the line, the angles BA P, A B P. being 54° and 30° respectively.

(4.) The perimeter of a triangle measures 10 inches, and two of its angles are 35° and 55° respectively. Construct it and measure the length of the sides. Ans.-2.4, 3·42, and 4·18 inches.

(5.) The sides of an equiangular hexagon are alternately 2 inches and 1 inch. Construct the figure and measure the lengths of its diagonals. Ans.-2.65 inches and 3 inches.

(6.) In a quadrilateral field, A B C D, the distance A D is 120 yards; B E and CF are perpendiculars upon A D; A E and D F are each 40 yards; B E measures 30 yards, and C F 60 yards. Construct a plan of the field, and give the measurements of the other sides and of the diagonals. Ans.-A B= 50 yards, B C=50 yards, C D=72·11 yards, A C=100 yards, B D=85·44 yards.

(7.) Describe a square upon a line 3 inches long; describe a triangle on the diagonal of the square as base, having its vertex on the middle point of one of the sides of the square; and measure the sides and angles of the triangle.—

Ans.-Sides, 1 inches, 5

inches, 44 inches. Angles, 18° 26', 45°, 116° 34'.

(8.) Find the height of a tower whose angle of elevation 200 yards off is 15°.-Ans. 53% yards.

(9.) A man finds that the angle of elevation of the top of a tower is 60°, he recedes from the tower 30 yards, and finds the angle reduced to 30°; find the height of the tower.-Ans. 26 yards.

(10.) A tower stands on a cliff; from a boat 100 yards from the base of the cliff, the elevation of the foot of the tower is 15°, the boat being moved 100 yards further off, the elevation of the top of the tower is 15°. Find the heights of the cliff and the tower.-Ans. Each 26 yards.

CHAPTER III.

ON THE DIVISION OF LINES AND THE CONSTRUCTION OF

SCALES.

THEOREMS.

103. THEOR. 15.—If a straight line be drawn parallel to one of the sides of a triangle it shall cut the other sides, or these produced, proportionally. (Eu. VI. 2.)

104. THEOR. 16.-In a right-angled triangle, if a perpen