364 
PAPPUS OF ALEXANDRIA 
the problem of representing the three means with the respective 
extremes by six lines drawn in a semicircle. 
Given a semicircle on the diameter AC, and B any point on 
the diameter, draw BD at right angles to AG. Let the tangent 
H V 
B c 
at D meet AC produced in G, and measure DH along the 
tangent equal to DG. Join MB meeting the radius OD in K. 
Let BF be perpendicular to OD. 
Then, exactly as above, it is shown that OK is a harmonic 
mean between OF and OD. Also BD is the geometric mean 
between AB, BC, while OC { — OD) is the arithmetic mean 
between AB, BC. 
Therefore the six lines DO (= OC), OK, OF, AB, BC, BD 
supply the three means with the respective extremes. 
But Pappus seems to have failed to observe that the ‘ certain 
other geometerwho has the same figure excluding the dotted 
lines, supplied the same in five lines. For he said that DF 
is ‘ a harmonic mean ’. It is in fact the harmonic mean 
between AB, BC, as is easily seen thus. 
Since ODB is a right-angled triangle, and BF perpendicular 
to OD, 
DF: BD = BD : DO, 
or DF. DO = BD 2 = AB. BC. 
But DO = \ (AB + BG); 
therefore DF. {AB + BC) = 2 AB . BC. 
Therefore AB. {DF— BC) = BC. {AB—DF), 
that is, AB : BC = {AB - DF): {DF- BC), 
and DF is the harmonic mean between AB, BC. 
Consequently the five lines DO {= OC), DF, AB, BC, BD 
exhibit all the three means with the extremes.