Problem 7.

To divide an Angle given into two equal Parts.

Let ABC be the given Angle, and let it be required to (bisect or) divide it into two equal Parts.

Construction. First, having opened the Compasses to any convenient Wideness, set one Foot in the Point B, and describe the Arch acb, cutting the Sides in a and b; then with the same or any Extent at Pleasure, setting one Foot in a, defcribe the Arch ff; with the fame Extent set one Foot in b, and describe the Arch ee, cutting the former at D. Lastly, draw the Line BD, and it will bisect, or divide the Angle into two equal Parts, as required.

By this Problem, a Quadrant may be divided into certain Degrees; and the Seaman's Compass expeditiously divided into 32 Points.

Problem 8.

To make an Angle equal to an Angle given.

Let ABC be the given Angle; and let it be required to make another Angle equal to it.

Conftruction. First, upon the Angular Point B, with any Opening of the Compasses, describe the Arch a b. Then, upon the Point E, (having drawn the Line FE) with the fame Extent of the Compaffes describe the Archcd; next, take the Arch ab in the Compasses, and set it from a to d. Lastly, draw the Line EdD, and it will make the Angle FED, equal to the Angle A B C, as was required.

* When an Angle is expressed by three Letters, the middle Letter expresses the Angular Point, and the first and laft the End of the Legs.

This Problem is so neceflary, that Surveying, Fortification, Perspective, &c. cannot be performed without it.

Problem 9.

To divide a given Right Line into any Number of equal Parts.

Let A B be the Line given, and let it be proposed to be divided into fix equal Parts.

Construction. First, from the End B of the given Line draw the Line BC, making an Angle of any Quantity with the Line AB. Then from the other End A draw the Line DA (parallel to BC by Problem 5, or) making the Angle DAB equal to the Angle CBA by the last Problem. Next, on the Line BC, with any small Opening of the Compafles, beginning at B, make the five equal Distances at 1, 2, 3, 4, 5. Also set off the fame Distances on the Line AD, beginning at the Point A. Lastly, draw Lines from 5 to 1; from 4 to 2; from 3 to 3, &c. as in the figure, and they will divide the given Line A B into 6 qual Parts, as required.

By this Problem, a Line may be divided after the ame Manner as another Line is divided, with great Exactefs and Precision.

Problem 10.

To make a Triangle, each Side of which shall be equal to a given Right Line.

Let L be the Line given, and let it be required to make a Triangle, having each Side equal to the faid Line.

Construction. First, draw the Line A C something longer than the given Line L. Take the Line L in your Compasses, and fet it from A to B. With the same Extent of the Compasses, place one Foot in A, and describe the Arch

a a.

Then place one Foot in B, and strike the Arch bb, intersecting the former in the Point D. Lastly, from D, draw Lines to A and to B, so will the Triangle be formed, having each Side equal to the Line L, as was required.

This Figure is called an Equilateral Triangle, because all its Sides are equal, and may be usefully applied (when described on a Board) in taking inaccessible Distances.

To make a Triangle whose three Sides shall be equal to three given Right Lines, provided that any two of them be greater than the third.

Let A, B, and C, be the three Lines given, with which it is required to make a Triangle.

Construction. First, take the Line A in the Compaffes, and set it from D to E; then take the Line B in the Compaffes, and setting one Foot in E, describe the Arch cc. That done, take the Line C in the Compaffes, and setting one Foot in D, describe the Arch e e, intersecting the other in F. Lastly, from F draw the Lines FD and FE, fo shall the Triangle be formed, whose three Sides are equal to three Lines, A, B, and C, respectively.

We make Use of this Problem in Surveying, &c. to make a Figure equal to another given, by dividing it into Triangles, and taking off the Sides as above.