If that's a true equation, the best way to prove it is with coordinate
geometry.



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 = 

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 = ,

with both sides squared:

d² = (x2 - x1)² + (y2 - y1)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 = 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