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 .