Lemma to Proposition 2.
[On Spirals, Prop. 10.]

If A1,  A2, A3,  ... A be n lines forming an ascending arithmetical progression in which the common difference is equal to the least term A1, then

(n+1)An2+A1(A1+A2+A3+...+An ) =3(A12+A22+A32 +... + An2)

Let the lines An, An-1, An-2 , ... A1 be placed in a row from left to right. Produce An-1 , An-2 , ...A1 until they are each equal to An , so that the parts produced are respectively equal to A1, A2, ... An-1 .

Taking each line successively, we have

And, by addition, (n+1)An2=2(A12 + A22 + ... + An2) +2A1 .An-1 + 2A2.An-2 +... +2An-1.A1.

Therefore, in order to obtain the required result, we have to prove that

2(A1 .An-1 + A2.An-2 +... +An-1.A1)+A1(A1 + A2 + ... + An)
= A12 + A22 + ... + An         ...... (*)
Now
2A2.An-2  = A1 .4An-2, because A2 = 2A1,
 2A3.An-3 = A1 .6An-3, because A3= 3A1,
 ................................
2An-1.A1=A1. 2(n - 1)A1.

It follows that
2(A1 .An-1 + A2.An-2 +... +An-1.A1)+A1(A1 + A2 + ... + An)

And this last expression can be proved to be equal to

A12 + A22 + ... + An2

For  An2 = A1 (n.An)

because (n-1)An = An-1+ A1
+ An-2+ A2
+ ...................
+ A1 +An-1

 
Sirnilarly An-12 = A1{An-1+ 2(An-2 + An-3 + ... + A1)}

A12 = A1.A1;

whence, by addition,
 A12 + A22 + ... + An2

= A1 {An + 3An-1 + 5An-2 ... + (2n - 1) A1}.

Thus the equation marked (*) above is true; and it follows that

(n+1)An2+A1(A1+A2+A3+...+An ) =3(A12+A22+A32 +... + An2)