Cor. From the above proportion, we have, that is, the square of the tangent is equal to the rectangle of the secant and its external segment. PRACTICAL APPLICATIONS. PROBLEM I. To aivide a given line into parts proportional to given lines, also into equal parts. 1°. Let AB be a given line, and let it be required to divide it into parts proportional to the lines P, Q and R. draw CI and DF parallel to EB: then will AI, IF, and FB, be proportional to P, Q, and R (P XV., C. 2). 2o. Let AH be a given line, and let it be required to divide it into any number of equal parts, say five. BH, and from I, K, L, and M, draw the lines IC, KD, LE, and MF, parallel to BH: then will AH be divided into equal parts at C, D, E, and F (P. XV., C. 2). PROBLEM II. To construct a fourth proportional to three given lines. to C; draw AC, and from B draw BX parallel to AC then will DX be the fourth proportional required. For (P. XV., C.), we have, or, DA : DB :: DC : DX A : B :: C : DX. Cor. If DC is made equal to DB, DX will be a third proportional to DA and DB, or to A and B. PROBLEM III. To construct a mean proportional between two given lines. then will EG be the mean proportional required. For (P. XXIII., C. 2), we have, To divide a given line into two such parts, that the greater part shall be a mean proportional between the whole line and the other part. it till it terminates in the concave arc at E; centre and AD as radius, describe the arc will AF be the greater part required. For, AB being perpendicular to CB at B, is tangent to the arc DBE: hence AE – AB : AB :: AB – AD : AD. Scholium. When a 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 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. B 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. CF G A E F Again, draw CE; produce AE and draw DF parallel to CE 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. |