THE METHOD 
31 
Now HA : AH — A'A : AN 
= KA : AQ 
= MN: NQ 
= ME 2 : MN. NQ. 
It is now necessary to prove that MN .NQ — NP 2 + NQ 2 . 
k ; a' k' 
By the property of the ellipse, 
AN. NA' :NP 2 = {\AA'f: 
= AN 2 : NQ 2 ; 
therefore i\ r Q 2 : NP 2 = AN 2 : AN. NA' 
= NQ 2 ; NQ. QM, 
whence NP 2 = MQ. QN. 
Add NQ 2 to each side, and we have 
NP 2 + NQ 2 = MN.NQ. 
Therefore, from above, 
HA:AN=MN 2 :(NP 2 + NQ 2 ). (1) 
But MN 2 , NP 2 , NQ 2 are to one another as the areas of the 
circles with MM', PP', QQ' respectively as diameters, and these