8. Construct a regular nonagon. SOLUTION.--First get a quarter of the circumference by marking the points where two diameters at right angles to each other would cut the circumference. AX is an arc of 90°. Then from A take AY = 60° by using radius as a chord. YX is therefore an arc of 30°. Divide this into three equal parts by trial. Measure YB equal to two-thirds of YX, and AB and BC are arcs of 40°, and the chords AB and BC are chords of the regular nonagon. 9. To draw a five-point star. SOLUTION.--Draw a circle, and dividing the circumference into five equal parts, join the alternate points of division, as in the figure. FIG. 101. 10. To circumscribe a square about a circle (56). Also an equilateral triangle, and a regular hexagon. *These are inserted simply to give completeness. Of course, the student is not expected to know more than their names. PART II. THE FUNDAMENTAL PROPOSITIONS OF ELEMENTARY GEOMETRY, DEMONSTRATED, ILLUSTRATED, AND APPLIED. CHAPTER I. PLANE GEOMETRY. SECTION I OF PERPENDICULAR STRAIGHT LINES. PROPOSITION I. 122. Theorem.-At any point in a straight line, one perpendicular can be erected to the line, and only one, which shall lie on the same side of the line. FIG. 102. DEM.-Let AB* represent any line, and P be any point therein; then, on the same side of AB there can be one and only one perpendicular erected at P. For from P draw any oblique line, as PC, forming with AB the two angles CPB and CPA. Now, while the extremity P, of PC, remains at P, conceive the line PC to revolve so as to increase the less of the two angles, as CPB, and decrease the greater, as CPA. It is evident that for a certain position of CP, as C'P, these * In class recitation the pupil should go to the blackboard, after having had his proposition assigned him, and first draw the figure required for the demonstration. This should be done neatly, accurately, with dispatch, and without any aids. The figure being complete, he stands at the board, pointer in hand, enunciates the proposition, and then gives the demonstration as it is in the text, pointing to the several parts of the figure as they are referred to. angles will become equal. In this position C'P becomes perpendicular to AB (26).* Again, if the line C'P revolve from the position in which the angles are equal, one angle will increase and the other diminish; hence there is only one position of the line on this side of AB in which the adjacent angles are equal. Therefore there can be one and only one perpendicular erected to AB at P, which shall lie on the same side of AB. Q. E. D. 123. COR. 1.-On the other side of the line a second perpendicular, and only one, can be drawn from the same point in the line. 124. COR. 2.-If one straight line meets another so as to make the angle on one side of it a right angle, the angle on the other side is also a right angle, and the first line is perpendicular to the second. 125. COR. 3.-If two lines intersect so as to make one of the four angles formed a right angle, the other three are right angles, and the lines are mutually perpendicular to each other. DEM. Thus, if CEB is a right angle, CEA, being equal to it, is also a right angle. Then, as AEC is a right angle, the adjacent angle AED is a right angle, since they are equal. Also, as CEB is a right angle, and BED equal to it, BED is a right angle. Hence CD being, perpendicular to AB, AB is perpendicular to CD, as it meets CD so as to make the adjacent angles AEC and AED, or CEB and BED equal to each other (43). A FIG. 103. B PROPOSITION II. 126. Theorem.-When two straight lines intersect at right angles, if the portion of the plane of the lines on one side of either line be conceived as revolved on that line as an axis until it coincides with the portion of the plane on the other A side, the parts of the second line will coincide. DI FIG. 104. DEM.-Let the two lines AB and CD intersect at right angles at E; and let the portion of the plane of the lines on the side of CD on which B lies be conceived to revolve on the line CD as an axis,† until it falls in the * When a preceding principle is referred to, it should be accurately quoted by the pupil. + As if the paper, which may represent the plane of the lines, were folded in the line CD. It is important that this process be clearly conceived, as it is to be made the basis of many subsequent demonstrations. portion of the plane on the other side of CD. Then will EB fall in and coincide with AE. For, the point E being in CD, does not change position in the revolution; and, as EB remains perpendicular to CD, it must coincide with EA after the revolution, or there would be two perpendiculars to CD on the same side and from the same point, E, which is impossible (122). Hence EB coincides with ÉA. Q. E. D. PROPOSITION III. 127. Theorem.-From any point without a straight line, one perpendicular can be let fall upon that line, and only one. DEM.-Let AB be any line, and P any point without the line; then one perpendicular, and only one, can be let fall from P upon AB. F DE CB For, conceive any oblique line, as PC, drawn, making the angle PCB>PCA. Now, while the extremity P of this line remains fixed, conceive the line to revolve so as to make the greater angle PCB decrease, and the less angle PCA increase. At some position of the revolving line, as PD, the two angles which it makes with the line AB will adjacent angles are equal, the line, as PD, is perMoreover, there is only one position of the line in which these angles are equal; hence, only one perpendicular can be drawn from a given point to a given line. Q. E. D. become equal. When these pendicular to AB (26, 43). PROPOSITION IV. 128. Theorem.-From a point without a straight line, a perpendicular is the shortest distance to the line. Α DEM.-Let AB be any straight line, P any point without it, PD a perpendicular, and PC any oblique line; then is PD <PC. P ם FIG. 106. Let the portion of the plane of the lines above AB be revolved upon AB as an axis, until it coincides with the portion below AB. Let P' be the point where P falls in the plane below AB. Now conceive the upper part of the plane as revolved back to its original position, and draw PP' and P'C. B Again, revolving the upper portion of the plane as before until P falls at P', since the points D and C remain fixed, the lines PD and P'D will coincide, as also the angles PDC and P'DC. Hence, PDC P'DC, and PD is the perpendicular from P upon AB (26,43,125). Moreover, PD = P'D and PC P'C, since they coincide when applied. Finally, PP' being a straight line, is shorter than |