A history of Greek mathematics - Wilbourhall.org

Now
THE METHOD
HA:AN=A'A:AN
= KA:AQ
= MN:NQ
= MN 2 :MN.NQ.
31
It is now necessary to prove that MN.NQ = NP 2 + NQ 2 .
H
*t
A
M
p/q/7 \\qNp' 0'
w \\\
BmN \ V B'
V \ ,
/e V
G C/ \F
M'
K A ; K'
By the property of the ellipse,
AN. NA' :
NP
2 = dAA') 2 : QBB')
2
= AN 2 :NQ 2 ;
therefore NQ 2 : NP 2 = AN 2 :AN. NA'
= NQ 2 :NQ.QM,
whence NP = MQ 2 . QN.
Add NQ 2
to each side, and we have
NP 2 + NQ 2 = MN.NQ.
Therefore, from above,
2
,
NQ
HA:AN= MN 2 : (NP 2 + NQ 2 ). (1)
But MN 2 , NP 2 are to one another as the areas of the
circles with MM', PP' ',
QQ' respectively as diameters, and these