Twitter is a great place to find geometry problems. The July 22, 2017 post of xylem presented the image below with two squares, *ABCD* and *BFGE,* sharing a vertex. Given that *AE* = 5, what is the length of *DG?*

My first thought was that surely the problem was underspecified. Without knowing more about the size of *ABCD* and the orientation of *BFGE,* how could the length of *DG* be unique? Well, there was one way to check—I built myself a Web Sketchpad model! The model below (and here on its own page) allows you to drag point *E,* keeping length *AE* constant (The length of *AE* is controlled by the horizontal segment to the left of the squares.)

Sure enough, regardless of point *E*‘s location, *DG*‘s length remains invariant. Before jumping to a proof, we can determine *DG*‘s length by picking special locations for point *E*. In the picture below left, point *E* lies on *AB.* We can see from the added dashed segments that *GD* is the diagonal of a square with side *AE,* so *GD* = 5√2. Another way to see the relationship between *AE* and *DG* is to drag point *E* onto *AB* and then zero-out square *BFGE* entirely by adjusting *AE’*s length until it equals *AB* (See the picture below right.)

To show that *GD* = 5√2 regardless of point *E*‘s location, Twitter users such as alaaddin çizer and Vincent Pantaloon offered a number of proofs. Below (and on page 2 of the websketch), we create parallelogram *AEGH*. Can you prove that Δ*GHD* is an isosceles right triangle?

As a second proof, the illustration below (and page 3 of the websketch) shows a blue square erected on *AE*. Can you prove that *EH* = *GD*?

Finally, one of the shortest proposed proofs requires only that you show Δ*GBD*∼Δ*EBA* with scale factor √2. I’ll leave the details to you.

The arrangement of the two squares in this problem reminded me of an area-based proof of the Law of Cosines first described to me by David Dennis. In the picture below (and on page 4 of the websketch), there are three squares and two congruent parallelograms. By sliding Δ*BFC* down onto Δ*AHD* and sliding Δ*CID* across onto Δ*BEA,* we can see that the sum of the areas of the orange square, the blue square, and the two parallelograms is equal to the area of square *ABCD*. The Law of Cosines activity from the Sketchpad module *Pythagoras Plugged In* by Dan Bennett challenges you to add the algebraic steps needed to show the connection to the familiar Law of Cosines statement. And as a bonus, if you drag point *E* so that ∠*BEA* = 90°, you create a geometric dissection proof of the Pythagorean Theorem.

If you flip through the various page of the websketch above, you’ll see that page 5 is left blank. You can use this page to create any of the interactive models from page 1-4 using the four tools that come with the websketch as well as our new format palette. The how-to movie below will get you started.