Using statement of Proposition 9 of Book II of Euclid's Elements:
"If a straight line is divided equally and also unequally, the sum of the squares on the two unequal parts is twice the sum of the squares on half the line and on the line between the points of section"
From this I have to obtain the following identity:
$$(a + b)^{2} + (a - b)^{2} = 2(a^2 + b ^2)$$
I have to show how this can be established geometrically and have done so for another identity in my review that was similar $a^2 - b^2 = (a +b)(a - b)$) Not sure if anything from that example would help but i have it for reference.
Any hints/ help with this is greatly appreciated.
Here is a very rough, first look, not too subtle geometric justification of this fact, as long as you are allowed to bring it to the configuration of Pythagoras' theorem, then you can apply one of the many geometric cut an paste ways to prove Pythagoras' theorem (I think Euclid knew about Pythagoras' theorem, or maybe he was the one who came up with it, I actually don't know this part of the story).
Look at the pic I have attached. $M$ is midpoint of $AB$, point $N$ is arbitrary. You have squares constructed to each appropriate segment. The two gray triangles are congruent because if you perform a $90^{\circ}$ clockwise rotation around point $Q$, triangle $AMQ$ is mapped to $QDP$, because $$AM = MB = MN + NB = MN + NP = DN + NP = DP$$ and $QM = QD$. Consequently, $AQ$ is mapped to $QP$ and thus $AQ$ is perpendicular to $PQ$ and $AQ = PQ$. Hence $APQ$ is an isosceles right angled triangle.
$$2(AM^2 + MN^2) = 2(AM^2 + MQ^2) = 2 \, AQ^2 =2 \, AP^2 = AQ^2 + AP^2 = AP^2$$ $$ AP^2 = AN^2 + NP^2 = AN^2 + NB^2$$ Hence $$2(AM^2 + MN^2) = AN^2 + NB^2$$