An interesting fact holds in a configuration of three equal squares placed side by side. If the lines drawn as shown then ∠BDC + ∠BFC = ∠BEC. There are many proofs of this result. These are so simple, the problem has been included in many a puzzle book.

First note that ∠BEC = 45°. The task is then to show that ∠BDC + ∠BFC = 45°. Let's introduce α = ∠BDC and β = ∠BFC. We are to prove that α + β = 45°.

Proof 1

Add another row of squares below the given three. ΔBKD is isosceles (BK = DK) and, in addition, ∠K = 90°. Therefore, ∠BDK = 45°. However, since triangles DEK and FCB are equal, we also have ∠KDE = β.

Proof 2

Now with some trigonometry (arctan = tan^-1),

α + β = arctan(1/3) + arctan(1/2)   = arctan((1/3 + 1/2)/(1 - 1/3·1/2))   = arctan(1),

where I used the standard formula for the tangent of a sum

Since arctan(1) = 45°, α + β = 45°.

Proof 3

Triangles BDE and FEB are similar. Indeed, assuming the side of the square equals 1, then

which implies that ∠BDC = ∠FBE = a. Finally, the exterior angle BEC of ΔFEB, equals the sum of the opposite interior angles, i.e., a + b.

In short, we proved that

Observe that this implies another identity:

Proof 4

Following a proof without words by Hasan Unal (Math Magazine, v 82, n 1, February 2009 , p. 56) we shall prove a more general statement. Let x and y be positive numbers satisfying x² + y² = 1. Then

arctan 1 - x y  +  arctan 1 - y x  =  π 4