Also, Lockhart's Lament is from 2002, so this post probably needs a (2002). It is very unlikely that the proof was new. It was certainly new to the seventh-grader, and a great result at that.
Thus, no parallelogram with non-right angles exists that has equal length diagonals.
My favorite book on the topic, unfortunately only in Slovak, is: "Matematici, ja a ty" (mathematicians, you and me). I don't think there is a translation. I also had a university professor who had a similar style - not dry proofs, but "Here is how Archimedes thought about proofs" - which is related to the comment above: if A is not smaller than B and B is not smaller than A, they need to be equal.
In my opinion it's a lot more obvious that a parallelogram with 2 equal diagonals must be a rectangle than that an inscribed angle intersecting a circle at ends of a diameter must be a right angle.
I think the two facts, by themselves, are roughly equivalent in difficulty. Moving from Thales to equal diagonals parallelogram requires drawing an extra circle; moving from parallelogram facts to Thales requires either the construction in the article, or continuing the missing radius to get the same rectangle.
Clearly, filling in all of the details of any formal proof in Greek style is going to require a careful knowledge of the axioms in use and some list of previously proven theorems which are allowed to be used without re-proof. Depending on which theorems are at hand already, one or another proof might be shorter or longer or more or less obvious. When trying to construct a whole mathematical treatise, the best order for the theorems so that each proof depends only on previously proven theorems is a tricky choice involving some trade-offs between pedagogical goals and concision of individual proofs.
But all of that is missing the point.
(But then you need to know that parallelogram diagonals split each other in half, …)
IIRC you prove it via angles.
a) The sum of angles of any triangle is pi.
b) The sum of angles of any quadrilateral is 2*pi.
c) Since you rotate the triangle, opposite angles end up adding together. Because of the above, a bit of reasoning around symmetry shows that when diagonals are equal they can only add to pi/2 and that the other ones can only be pi/2. Any other angle leads to a contradiction.
I mentioned angle values but (again IIRC) this can all be proven with compass and ruler.
But then, the article is about laying out a proof of Thales so if it's using a lemma that leveraged Thales itself that's circular logic right there and so we're in trouble.
One other way to prove Thales is by computing angles, but then you've got yourself a proven Thales so the point of the article is moot.
By the point you've proven Thales with isosceles (b), proving that a parallelogram with equal diagonals is a rectangle thanks to Thales(b) in order to go forward along the Thales(a) proof is circular logic.
The net change in direction around a triangle, quadrilateral, or any simple closed curve is a full rotation, i.e., 2π.
The sum of the interior angle and the signed change in direction around a vertex of a polygon is π (consider the case of a "vertex" with interior angle π).
So for a polygon with interior angles θ₁, θ₂, …, θₙ,
2π = (π - θ₁) + (π - θ₂) + ⋯ + (π - θₙ),
therefore
θ₁ + θ₂ + ⋯ + θₙ = nπ - 2π = (n-2)π.