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. |