An Elementary Proof of Bottema's Theorem

Proof

Let's drop perpendiculars B_aU, MW, A_bV, and CZ onto AB.

MW is the midline of trapezoid B_aUVA_b so that

MW = (B_aU + A_bV) / 2.

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

B_aU = AZ.

Similarly,

A_bV = BZ.

Taking all three identities into account shows that

MW = (B_aU + A_bV) / 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.)

Reference

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

Bottema's Theorem

1. Bottema's Theorem
2. An Elementary Proof of Bottema's Theorem
3. On Bottema's Shoulders
4. On Bottema's Shoulders II
5. Friendly Kiepert's Perspectors
6. Bottema Shatters Japan's Seclusion
7. Rotations in Disguise
8. Four Hinged Squares
9. Four Hinged Squares, Solution with Complex Numbers
10. Pythagoras' from Bottema's

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