An Elementary Proof of Bottema's Theorem

Consider a configuration of two squares ACBcBa and BCAcAb with a common vertex C.

Bottema's theorem

Bottema's theorem claims that the midpoint M of the segment AbBa is independent of C.

Proof

Let's drop perpendiculars BaU, MW, AbV, and CZ onto AB.

An Elementary Proof of Bottema's    Theorem

MW is the midline of trapezoid BaUVAb so that

MW = (BaU + AbV) / 2.

Further, since ∠BaAC is right, angles BaAU and CAZ are complementary which makes right triangles BaAU and ACZ equal, implying

BaU = AZ.

Similarly,

AbV = BZ.

Taking all three identities into account shows that

MW = (BaU + AbV) / 2 = (AZ + BZ) / 2 = AB/2 = AW,

independent of C. (The proof also shows that M is the center of the square formed on AB upwards.)

Note: there is a dynamic illustration of the proof that makes it entirely obvious.

Reference

  1. A. Shriki, Back To Treasure Island, MathematicS Teacher, Vol. 104, No. 9 (May 2011), 658-664
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