PART I. Problems resolved by help of the Ruler, or by PROP. I. PROB. To bisect a given angle. Let BAC be an angle, which it is required to bisect, by drawing only straight lines. In AB take any two points D and E, from AC cut off AF equal to AD and AG to AE, draw EF and DG, crossing in the point H: AH will bisect the angle BAС. For the triangles EAF and DAG, having the sides EA and AF equal by construc triangles DEH and HGF, having the angle at E equal to that at G, the vertical angles at H equal, and also their opposite sides DE and FG, are equal (I. 20.); and hence the side DH is equal to FH. Again, the sides AD and DH are equal to AF and FH, and AH is common to the two triangles AHD and AHF, which are therefore equal (I. 2.), and consequently the angle DAH is equal to FAH. PROP. II. PROB. To bisect a given finite straight line. Let it be required to bisect AB, by a rectilineal construction. Draw AK diverging from AB, and make AC=CD=DE, join EB, and continue it beyond B till BF be equal to BE, and lastly join FC; which will bisect AB in the point G. For draw BH parallel to AE. And because BD evidently bisects the sides EC and EF of the triangle CEF, it is parallel to the base CF (VI. 1. cor. 2.); wherefore BDCH is a parallelogram, which has (I. 26.) its opposite sides BH and CD equal. But AC being parallel to BH, the angles GAC and GCA are equal to GBH and GHB, and F H A B G C D E the side AC, being made equal to CD, is hence equal to its cor K responding interjacent side BH; whence the triangles AGC and BGH are equal (I. 20.), and therefore AG is equal to BG. f PROP. III. PROB. Through a given point, to draw a line parallel to a given straight line. Let it be required, by a rectilineal construction, to draw through C a straight line parallel to AB. C G In AB take any two points D and F, join CD, which produce till DE be equal to it; again join E with the point F, and continue this till FG be equal to EF: Then CG, being joined, will be parallel to AB. For, since AB or DF evident ly bisects the sides EC and EG A D FB E of the triangle CEG, it must be parallel to the base CG (VI. 1. cor. 2.). PROP. IV. PROB. From a point in a given straight line, to erect a perpendicular. Let C be a given point, from which it is required, by help of straight lines merely, to erect a perpendicular to AB. In AB, having taken any point D, draw DE equal to DC and inclined to AB, join EC and produce it until CG be equal to CD or DE, make CF equal to CE, join FG and produce this till GH be equal to GC: Then CH will is a right angle (III. 19.), or CH is perpendicular to AB. PROP. V. PROB. To let fall a perpendicular upon a given straight line, from a point without it. Let C be a given point, from which it is required, by a rectilineal construction, to let fall a perpendicular to AB. In AB take any point D, draw DF obliquely, and make DE =DF=DG, join FE and produce it until CK parallel to it: CK is the perpendicular required. For the point D being obviously the centre of a semicircle passing through G, F, and E, the angle GFE is a right angle; and the triangles EGF, EHI, having the sides GE, EF equal to HE, EI, and their contained angles vertical, are equal (I. 3.), and consequently the angle HIE is equal to GFE, or is a right angle; but since CK and HI are parallel, the angle CKA is equal to HIE (I. 22.), and therefore is also a right angle, or CK is perpendicular to AB. PART II. Geometrical Problems resolved by means of Compasses, or by the mere description of Circles. PROP. I. PROB. To repeat a given distance in the same direc tion. Let A and B be two given points; it is required to find, by means of compasses only, a series of equidistant points in the same extended line. From B as a centre, with the given distance BA, describe a portion of a circle, in which inflect that distance three times to C; from C, with the same radius, describe |