# An Elementary Proof of Bottema's Theorem

Consider a configuration of two squares ACB_{c}B_{a} and BCA_{c}A_{b} with a common vertex C.

*Bottema's theorem* claims that the midpoint M of the segment A_{b}B_{a} is independent of C.

### Proof

Let's drop perpendiculars B_{a}U, MW, A_{b}V, and CZ onto AB.

MW is the midline of trapezoid B_{a}UVA_{b} so that

MW = (B_{a}U + A_{b}V) / 2.

Further, since ∠B_{a}AC is right, angles B_{a}AU and CAZ are complementary which makes right triangles B_{a}AU and ACZ equal, implying

B_{a}U = AZ.

Similarly,

A_{b}V = BZ.

Taking all three identities into account shows that

MW = (B_{a}U + A_{b}V) / 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

- A. Shriki,
__Back To Treasure Island__,*MathematicS Teacher*, Vol. 104, No. 9 (May 2011), 658-664

### Bottema's Theorem

- Bottema's Theorem
- An Elementary Proof of Bottema's Theorem
- Bottema's Theorem - Proof Without Words
- On Bottema's Shoulders
- On Bottema's Shoulders II
- On Bottema's Shoulders with a Ladder
- Friendly Kiepert's Perspectors
- Bottema Shatters Japan's Seclusion
- Rotations in Disguise
- Four Hinged Squares
- Four Hinged Squares, Solution with Complex Numbers
- Pythagoras' from Bottema's
- A Degenerate Case of Bottema's Configuration
- Properties of Flank Triangles
- Analytic Proof of Bottema's Theorem
- Yet Another Generalization of Bottema's Theorem
- Bottema with a Product of Rotations
- Bottema with Similar Triangles
- Bottema in Three Rotations
- Bottema's Point Sibling

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny67254548