Problems. PROBLEMS RELATING TO THE FOURTH BOOK. PROBLEM I. To divide a line into any proposed number of equal parts Let AB be the line, and let it be required to divide it into four equal parts. Draw any other line, AC, forming an angle with AB, and take any dis E D tance, as AD, and lay it off four times on AC. Join C and B, and through the points D, E, and F, draw parallels to CB These parallels to BC will divide the line AB into parts proportional to the divisions on AC (Th. xiii): that is, into equal parts. To find a third proportional to two given lines. A Let A and B be the given lines. AC, making an angle with it. On BC. then lay off AD, also equal to . B, and through D draw DE parallel to BC: then will AE be the third proportional sought. For, since DE is parallel to BC, we have (Th. xiii) AB AC :: AD or AC : AE; therefore, AE is the third proportional sought Problems. PROBLEM III. A To find a fourth proportional to the lines A, B, and C. Place two of the lines forming an angle with each other at A; that is, Bmake AB equal to A, and AC equal B; also, lay off AD equal to C. Then join BC, and through D draw C E DE parallel to BC, and AE will be the fourth proportional sought. For, since DE is parallel to BC, we have AB : AC :: AD : AE; therefore, AE is the fourth proportional sought. PROBLEM IV. To find a meun proportional between two given lines, A and B. Make AB equal to A, and BC equal to B: on AC describe a semicircle. Through B draw BE perpendicular to A B AC, and it will be the mean proportional sought (Th. xviii. Cor). PROBLEM V. To make a square which shall be equivalent to the sum of two given squares. Let A and B be the sides of the given squares. Draw an indefinite line AB, and make AB equal to A. At B draw BC perpendicular to AB, and make. BC equal to B: then draw AC and the square described on AC will be equivalent to the squares on A and B (Th. xii). Problems. PROBLEM VI. To make a square which shall be equivalent to the difference be tween two given squares. Let A and B be the sides of the given squares. Draw an indefinite line, and make CB equal to A, and CD equal to B. At D draw DE B perpendicular to CB, and with C as a centre, and CB as a radius, describe a semicircle meeting DE in E, and join CE: then will the square described on ED be equal to the difference between the given squares. For, CE is equal to CB, that is, equal to A, and CD is equal to B: and by (Th. xii. Cor.), ED-CE2-CD. PROBLEM VII. To make a triangle which shall be equivalent to a given quadrilateral. Let ABCD be the given quadri lateral. Draw the diagonal AC, and through D draw DE parallel to AC, meeting A BA produced at E. Join EC: then will the triangle CEB be equivalent to the quadrilateral BD. For, the two triangles ACE and ADC, having the same base AC, and the vertices of the angles D and E in the same line DE parallel to AC, are equivalent (Th. ii). If to each, we add ACB, we shall then have the triangle ECB equivalent to the quadrilateral BD (Ax. 2). To make a triangle which shall be equivalent to a given polygon Let ABCDE be the polygon. Draw the diagonals AD, BD. Produce AB in both directions, and through C and E draw CG and EF, respectively parallel to AD and BD: then join FD and D DG, and the triangle FDG will be equivalent to the polygon ABCDE. For, the triangle AED is equivalent to the triangle AFD, and DBC to DBG (Th. ii); and by adding ADB to the equals. we shall have the triangle FDG equivalent to the polygon ABCDE. PROBLEM IX. To make a rectangle that shall be equivalent to a given triangle Let ABC be the given triangle. Bisect the base AB at D, and draw DH perpendicular to AB. Through C, the vertex of the triangle, draw CHG parallel to AB, and draw BG perpendicular to it then will the rectangle DG be equivalent to the triangle ABC. CH For, the triangle would be half a rectangle having the same. base and altitude: hence, it is equivalent to DG, whose base is the half of AB, and altitude equal to that of the triangle. For, draw OA. Then in the two right angled triangles OAG and OAF, the side AO is common, and AG is equal to AF, since each is half of one of the equal sides of the polygon: hence, OG is equal to OF (Bk. I.Th. xix). In the same manner it may be shown that OH, OK and OL are all equal to each other hence, a circle described with the centre O and radius OF will be inscribed in the polygon. : Cor. Hence, also the lines OA, ON &c., drawn to the angles of the polygon are equal. APPENDIX OF THE REGULAR POLYGONS. 1. In a regular polygon the angles are all equal to each other (Def. 3). If then, the sum of the inward angles of a regular polygon be divided by the number of angles, the quotient will be the value of one of the angles. But the sum of the inward angles is equal to twice as many right angles, wanting four, as the polygon has sides, and we shall find the value in degrees by simply placing 90° for the right angle. |