Fun trapezoid coincidence on twitter tonight

Earlier tonight I saw this great question on twitter from Wendy Menard:

Math Club is stuck on this problem http://t.co/ZotjAlmrVb – is it possible to solve w/o trig? #MTBoS pic.twitter.com/RTe0p9J5jU

— Wendy Menard (@wmukluk) January 14, 2015

Dave Radcliffe posted a link to his really clever solution:

@Five_Triangles Since the triangles ACD and BCD have equal areas, the regions ABC and ABD also have equal areas. pic.twitter.com/KYOVfYbiGA

— Dave Radcliffe (@daveinstpaul) December 19, 2014

The discussion following Dave’s tweet was about why the triangles ACD and BCD in his picture have the same height (their areas are the same because the have the same base and the same height, but as Dave says in the discussion, the fact that the heights are the same isn’t obvious).

Well . . . funny enough before I saw Wendy Menard’s post tonight my son and I were talking through problem 20 from the 2008 AMC 10a:

Problem #20 from the 2008 AMC 10 A

Here’s the problem: Trapezoid ABCD has bases and and diagonals intersecting at K. Suppose that AB = 9, DC = 12, and the area of is 24. What is the area of trapezoid ABCD?

Working through the AMC 10 problem you’ll see that triangles AKD and BKC have the same area. That fact gives a little insight into why the two shaded triangles in Dave’s picture have the same area (and why the heights of triangles ABD and BCD are the same) -> ABCD in Dave’s picture is a trapezoid because angle DAB is 72 degrees!

How fun that two problems brought to our attention almost randomly have really similar ideas that help you get to the solution!

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