Dynamic Geometry Software. Step-by-Step construction, Manipulation, and animation

Draw squares ABDE and BCFG on sides AB and BC of a triangle ABC. Then the midpoint M of EF is independent of B and the triangle AMC is an isosceles right triangle.

Continue reading at:

gogeometry.com/geometry/bottema_theorem_triangle_square.htm

## Monday, December 22, 2008

### Bottema's Theorem: Triangle and Squares

Subscribe to:
Post Comments (Atom)

this's actually not hard to prove.

ReplyDeleteT: midpoint of AC

Let M,N,P be the points so that EM,FN, BP is perpendicular to AC.

easily have: EM+FN=AC=2MT => MT // EM=> M is independant