will their segments be reciprocally proportional; that is, one' segment of the first will be to one segment of the second, as the remaining segment of the second is to the remaining segment of the first. For, draw CA and BD. Then will the angles ODB and OAC be equal, because each is measured by half of the arc CB (B. III., P. XVIII.). The angles OBD and OCA, will also be equal, because each is measured by half of the arc AD: hence, the triangles OBD and OCA are similar (P. XVIII., C.), and consequently, their homologous sides are proportional: hence, DO AO :: OB: OC; which was to be proved. Cor. From the above proportion, we have, DO X OC = AO × OB; that is, the rectangle of the segments of one chord is equal to the rectangle of the segments of the other. If from a point without a circle, two secants be drawn ter minating in the concave arc, they will be reciprocally proportional to their external segments. Let OB and OC be two secants terminating in the concave arc of the circle BCD: then will OB : OC :: OD : OA. For, draw AC and DB. The triangles ODB and OAC have the angle O common, and the angles OBD and OCA equal, because each is measured by half of the arc AD: hence, they are similar, and consequently, their homologous sides are proportional; whence, that is, the rectangles of each secant and its external segment are equal. If from a point without a circle, a tangent and a secant be drawn, the secant terminating in the concave arc, the tangent will be a mean proportional between the secant and its external segment. Let ADC be a circle, OC a secant, and OA a tangent: then will OC : ОА :: OA OD. For, draw AD and AC. The triangles OAD and OAC will have the angle common, and the angles OAD and ACD equal, because each is measured by half of the arc AD (B. III., P. XVIII., P. XXI.); the triangles are therefore similar, and consequently, their homologous sides are proportional: hence, OC : OA :: OA : OD; which was to be proved. 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 divide a given straight line into parts proportional to given straight lines: also into equal parts. 1o. Let AB be a given straight 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 straight 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 straight 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, DA : ᎠᏴ :: DC: DX; or, A : B C: DX. Cor. If DC is made equal to DB, DX will be third proportional to DA and DB, or to A and B. PROBLEM III. To construct a mean proportional between two given straight lines. Let A and B be the given lines. On an indefinite line, lay off DE equal to A, and EF equal to B; on DF as a diameter describe the semi-circle DGF and draw EG perpendicular to DF: D E Br AH then will EG be the mean proportional required. or, For (P. XXIII., C. 2), we have, DE : EG :: EG: EF; A: EG :: EG : B. PROBLEM IV. To divide a given straight line into two such parts, that the greater part shall be a mean proportional between the whole line and the other part. Let AB be the given line. At the extremity B, draw BC perpendicular to AB, and make it equal to half of AB. With C as a centre, and CB as a radius, describe the are DBE; draw AC, and produce E; it till it terminates in the concave arc at centre and AD as radius, describe the arc will AF be the greater part required. D with A as DF: then |