Question 622387
<pre>
If that's a true equation, the best way to prove it is with coordinate
geometry.

{{{drawing(300,250,-1,5,-1,4, line(0,-2,0,5),line(-2,0,5,0),
circle(2,3/4,.05), circle(1,3/4,.05), circle(1,3/2,.05), 
 locate(4,0,A), locate(0-.25,3+.2,B), locate(-.25,0,C),
locate(0-.25,3/2+.2,D), locate(2,3/2+.3,E), locate(2,0,F),
locate(2,3/4,P), locate(1,3/4,Q),locate(1,3/2+.3,R),

line(0,3,4,0), green(line(0,3,2,0),line(0,0,2,3/2), line(0,3/2,4,0)) )}}}

But the sum of all those squares will obviously be greater than 
just AB², but I'll try to prove it anyway, and see what happens.

Suppose A has the coordinates A(4a,0) and B has coordinates B(0,4b).

[I chose 4a and 4b instead of a and b to avoid fractions with the midpoint formula)

Then we use the midpoint formula,

Midpoint = {{{(matrix(1,3,      (x[1]+x[2])/2,   ",", (y[1]+y[2])/2))}}}

to find all these coordinates:

A(4a,0)
B(0,4b)
C(0,0)
D(0,2b)
E(2a,2b)
F(2a,0)
P(2a,b)
Q(a,b)
R(a,2b)

Then we use the distance formula, which is

d = {{{sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}},

with both sides squared:

d² = (x<sub>2</sub> - x<sub>1</sub>)² + (y<sub>2</sub> - y<sub>1</sub>)²

to find

PA² = (4a-2a)²+(0-b)² = (2a)²+(-b)²  = 4a²+ b²
PB² = (0-2a)²+(4b-b)² = (-2a)²+(3b)² = 4a²+9b²
PC² = (0-2a)²+(0-b)² = (-2a)²+(-b)²  = 4a²+ b²
QA² = (4a-a)²+(0-b)² = (3a)²+(-b)²   = 9a²+ b²
QB² = (0-a)²+(4b-b)² = (-a)²+(3b)²   =  a²+9b² 
QC² = (0-a)²+(0-b)² = (-a)²+(-b)²    =  a²+ b²
RA² = (4a-a)²+(0-2b)² = (3a)²+(-2b)² = 9a²+4b²
RB² = (0-a)²+(4b-2b)² = (-a)²+(2b)²  =  a²+4b²
RC² = (0-a)²+(0-2b)² = (-a)²+(-2b)²  =  a²+4b²
-----------------------------------------------
 PA²+PB²+PC²+QA²+QB²+QC²+RA²+RB²+RC² =34a²+34b²

But

AB² = (0-4a)²+(4b-0)² = (-4a)²+(4b)² = 16a²+16b²

So the equation you were asked to prove is not true.

What is true is

 PA²+PB²+PC²+QA²+QB²+QC²+RA²+RB²+RC² = {{{17/8}}}AB²

Sorry, but what you were asked to prove just isn't true.

Were some of those plus signs supposed to be minus signs?

Edwin</pre>