The Theorem of Apollonius (also known as Stewart's
Theorem) states that in any triangle ABC as shown,
m  a 2  n  b2  c  CM 2  m  n2  n  m2
C
b
a
I've never known this theorem, but it seems amazingly useful for
calculating the length CM of a median of a triangle (if M is the midpoint
of AB) based only on the lengths of the sides of the triangle.
c
A
m
n
M
B
c=m+n
Using the Theorem of Apollonius and AA' B' . We
have the following derivation:
e  j 2  f  a 2  A' B'r 2  e  f 2  f  e 2
2
d 2 d 2
d d
d d
 j  a  d r2    
2
2
2 4
2 4
2
d 2
d
d
( j  a 2 )  ( 2r 2  )
2
2
2
2
d
d2
j 2  a 2  2r 2 
, but
 a2
2
2
2
2
so j  2r and j  r 2
B'
2
j
f
a
M
r
A
s
d
B
e
k
a
Similarly, k  2 s and k  s 2
2
2
A'
In the figure above,
A' B'  d  a 2
To finish the proof that point M lies on a lemniscate, notice that AA'BB' is an isosceles trapezoid
(the proof is easy and left to the reader - look at congruent triangles within the figure and
discover congruent angles that show j || k)
Ptolemy's Theorem states that in a cyclic quadrilateral (such as AA'BB') the sum of the products
of opposite sides equals the product of the diagonals. So we have:
j  k  a  a  d (e  f )
j  k  a2  d 2
r 2  s 2  a 2  2a 2
rs 
a2
2
Since r  s is a constant, M must lie on a lemniscate. Q.E.D.
I am grateful to Lockwood (1961) for hints to use Apollonius' and Ptolemy's Theorems for this proof.
Lockwood, E.H. (1961). A Book of Curves. (p116). London: Cambridge University Press.