For, AB being perpendicular to CB at B, is tan. gent to the arc DBE: hence Scholium. When a straight line is divided so that the greater segment is a mean proportional between the whole line and the less segment, it is said to be divided in extreme and mean ratio. Since AB and DE are equal, the line AE is divided in extreme and mean ratio at D; for we have, from the first of the above proportions, by substitution, AE DE :: DE : AD. PROBLEM V. Through a given point, in a given angle, to draw a straight line so that the segments between the point and the sides of the angle shall be equal. Let BCD be the given angle, and A the given point. Through A, draw AE parallel to but, FE is equal to EC; hence, FA is equal to AD. PROBLEM VI. To construct a triangle equal to a given polygon. Let ABCDE be the given polygon. Draw CA; produce EA, and draw BG parallel to CA; draw the line CG. Then the triangles BAC and GAC have the common base AC, and because their vertices B and G lie in the same line BG parallel to the base, their altitudes are equal, and consequently, the triangles are equal: hence, the polygon GCDE is equal to the polygon ABCDE. A E Again, draw CE; produce AE and draw DF parallel to CE; draw also CF; then will the triangles FCE and DCE be equal: hence, the triangle GCF is equal to the polygon GCDE, and consequently, to the given polygon. In like manner, a triangle may be constructed equal to any other given polygon. PROBLEM VII. To construct a square equal to a given triangle. Let ABC be the given triangle, AD its altitude, and BC its base. Construct a mean proportional between AD and half of BC (Prob. III.). Let XY be that mean proportional, and on B it, as a side, construct a A X square then will this be the square required. For, from the construction, XY2 = {BC × AD = area ABC. Scholium. By means of Problems VI. and VII., a square may be constructed equal to any given polygon. PROBLEM VIII. On a given straight line, to construct a polygon similar to a given polygon. Let FG be the given line, and ABCDE the given polygon. Draw AC and AD. In like manner, construct the triangle FHI similar to ACD, and FIK similar to ADE; then will the polygon FGHIK be similar to the polygon ABCDE (P. XXVI., C. 3). PROBLEM IX. To construct a square equal to the sum of two given a square equal to the difference of two squares: also given squares. 1o. Let A and B be the sides of the given squares, construct a square: this square will be equal to the sum D a radius, describe an arc cutting DE at Scholium. A polygon may be constructed similar to either of two given polygons, and equal to their sum or difference. For, let A and B be homologous sides of the given polygons Find a square equal to the sum or difference of the squares on A and B; and let X be a side of that square. On X as a side, homologous to A or B, construct a polygon similar to the given polygons, and it will be equal to their sum or difference (P. XXVII., C. 2). lectoler 4.1 gm 5 Jure All Sept 16. 1911 BOOK V. REGULAR POLYGONS.-AREA OF THE CIRCLE. DEFINITION. 1. A REGULAR POLYGON is a polygon which is both equilateral and equiangular. PROPOSITION I. THEOREM. Regular polygons of the same number of sides are similar. Let ABCDEF and abcdef be regular polygons of the same number of sides: then will they be similar. angles, divided by the number of angles (B. I., P. XXVI, C. 4); and further, the corresponding sides are proportional, because all the sides of either polygon are equal (D. 1): hence, the polygons are similar (B. IV., D. 1); which was to be proved. |