The Math Teachers at Play blog carnival had several articles that especially drew my attention.
It requires hardly an explanation and beautifully applies to explain the division of fractions process:
It is a pity the Wikiversity page offered no references.
John Chase took an issue with the "exclusive" definition of trapezoid, as a quadrilateral that has exactly one pair of parallel sides. The inclusive definition, only stipulates the existence of such a pair, not necessarily uniqueness. According to the inclusive definition, a parallelogram is also a trapezoid, while the exclusive definition would not allow that. (There is interactive classification of quadrilaterals, with further discussion on the distinction between two definitions.
My own preference is unequivocally in favor of inclusive view. However, I remember having a similar discussion years ago where one participant found an example where parallelograms would not fit nicely. Unfortunately, I could not find my notes now.
Pat Ballew discussed the problem of reconstructing a triangle from the midpoints of its sides. He showed a simple vector solution and wondered why vectors are "almost totally ignored in US Geometry classes...".
The reason I think is that the problems could be classified by topics and also by the methods of solution. There is an interesting book Introduction to The Design & Analysis of Algorithms by A. Levitin that goes against the customary arrangement of such books by problems which are solved algorithmically. Levitin's book instead is arranged by algorithms which are used to solve various kinds of problems.
When problems are classified by topics a solution is expected from within that topic. Vector algebra solves many problems with ease, but, I guess, school curriculum just does not have room enough to cover everything that could be covered. Choices have to be made.
That particular problem of reconstructing a triangle from its midpoints admits a simple synthetic solution. Let the midpoints be L, M, N. Find the midpoints A', B', C' of ΔLMN. Through the vertices L, M, N draw the lines parallel to the sides of ΔA'B'C'. The three line form the solution to the problem. How simple that could be?
Also, the problem is nicely expandable to more general polygons.