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.
Dem.—Through E draw EM parallel to BC; make EN = EM; produce AB until EP = AB. Through the points N, P draw NT, RP each parallel to AC, and through P draw PQ parallel to BC. It is easy to see from the figure that the parallelogram QR is equal to the parallelogram MF, and is therefore given. Through P draw ST perpendicular to EP. Now, since AE = DB, BP is equal to DB; therefore PS = CD. Again, since OP = AD, PT is equal to CD; therefore PS = PT. Hence QR is the maximum parallelogram in the triangle SV T.