Another solution can be inferred from Proposition viii. For if AD, DC in that Proposition be respectively equal to X and Y , then DB will be the third proportional. Or again, if in the diagram, Proposition viii., AD = X, and AC = Y , AB will be the third proportional. Hence may be inferred a method of continuing the proportion to any number of terms. Exercises. 1. If AOΩ be a triangle, having the side AΩ greater than AO; then if we cut off AB = AO, draw BB′ parallel to AO, cut off BC = BB′, &c., the series of lines AB, BC, CD, &c., are in continual proportion. 2. AB − BC : AB :: AB : AΩ. This is evident by drawing through B′ a parallel to AΩ. PROP. XII.—Problem. To find a fourth proportional to three given lines (X, Y, Z). Sol.—Draw any two lines AC, AE, making an angle; then cut off AB equal X, BC equal Y , AD equal Z. Join BD, and draw CE parallel to BD. DE will be the fourth proportional required. Dem.—Since BD is parallel to CE, we have [ii.] AB : BC :: AD : DE; therefore X : Y :: Z : DE. Hence DE is a fourth proportional to X, Y , Z. Or thus: Take two lines AD, BC intersecting in O. Make OA = X, OB = Y , OC = Z, and describe a circle through the points A, B, C [IV. v.] cutting AD in D. OD will be the fourth proportional required. The demonstration is evident from the similarity of the triangles AOB and COD. PROP. XIII.—Problem. To find a mean proportional between two given lines. (X, Y ). Sol.—Take on any line AC parts AB, BC respectively equal to X, Y . On AC describe a semicircle ADC. Erect BD at right angles to AC, meeting the semicircle in D. BD will be the mean proportional required. Dem.—Join AD, DC. Since ADC is a semicircle, the angle ADC is right [III. xxxi.]. Hence, since ADC is a right-angled triangle, and BD a perpendicular from the right angle on the hypotenuse, BD is a mean proportional [viii. Cor. 1] between AB, BC; that is, BD is a mean proportional between X and Y . Exercises. 1. Another solution may be inferred from Proposition viii., Cor. 2. 2. If through any point within a circle the chord be drawn, which is bisected in that point, its half is a mean proportional between the segments of any other chord passing through the same point. 3. The tangent to a circle from any external point is a mean proportional between the segments of any secant passing through the same point. 4. If through the middle point C of any arc of a circle any secant be drawn cutting the chord of the arc in D, and the circle again in E, the chord of half the arc is a mean proportional between CD and CE. 5. If a circle be described touching another circle internally and two parallel chords, the perpendicular from the centre of the former on the diameter of the latter, which bisects the chords, is a mean proportional between the two extremes of the three segments into which the diameter is divided by the chords.