The problem to find two mean proportionals is one of the most celebrated in Geometry on account of the importance which the ancients attached to it. It cannot be solved by the line and circle, but is very easy by Conic Sections. The following is a mechanical construction by the Ruler and Compass. Sol.—Let the extremes AB, BC be placed at right angles to each other; complete the rectangle ABCD, and describe a circle about it. Produce DA, DC, and let a graduated ruler be made to revolve round the point B, and so adjusted that BE shall be equal to GF; then AF, CE are two mean proportionals between AB, BC. Dem.—Since BE is equal to GF, the rectangle BE.GE = BF.GF. Therefore DE.CE = DF.AF; hence DE : DF :: AF : CE; and by similar triangles, AB : AF :: DE : DF, and CE : CB :: DE : DF. Hence AB : AF :: AF : CE; and AF : CE :: CE : CB. Therefore AB, AF, CE, CB are continual proportions. Hence [VI. Def. iv.] AF, CE are two mean proportionals between AB and BC. The foregoing elegant construction is due to the ancient Geometer Philo of Byzantium. If we join DG it will be perpendicular to EF. The line EF is called Philo’s Line; it possesses the remarkable property of being the minimum line through the point B between the fixed lines DE, DF. Newton’s Construction.—Let AB and L be the two given lines of which AB is the greater. Bisect AB in C. With A as centre and AC as radius, describe a circle, and in it place the chord CD equal to the second line L. Join BD, and draw by trial through A a line meeting BD, CD produced in the points E, F, so that the intercept EF will be equal to the radius of the circle. DE and FA are the mean proportionals required. Dem.—Join AD. Since the line BF cuts the sides of the △ ACE, we have AB.CD.EF = CB.DE.FA; but EF = CB; therefore AB.CD = DE.FA, or = . Again, since the △ ACD is isosceles, we have ED.EC = EA2 − AC2 = (FA + AC)2 − AC2 = 2FA.AC + FA2 = FA.AB + FA2. Hence ED(ED + CD) = FA(AB + FA), or DE2 = FA.AB, therefore DE2 = FA.AB, and we have AB.CD = DE.FA. Hence AB, DE, FA, CD are in continued proportion. ________________ NOTE E. on philo’s line. I am indebted to Professor Galbraith for the following proof of the minimum property of Philo’s Line. It is due to the late Professor Mac Cullagh:—Let AC, CB be two given lines, E a fixed point, CD a perpendicular on AB; it is required to prove, if AE is equal to DB, that AB is a minimum.