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^o. The task is then to show that BDC + BFC = 45^o. Let's introduce a = BDC and b = BFC. We are to prove that a + b = 45^o.

Proof 1

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

Proof 2

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

a + b = 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^o, a + b = 45^o.

Proof 3

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

DE = 2, BE = 2, DE/BE = 2, BE = 2, EF = 1, BE/EF = 2, BD = 10, BF = 5, BD/BF = 2,

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

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