Hi thereddevils,
Draw parallelogram ABCD with AB being the top side. Extend AB to E such that BE = DC.
Connect D to B and C to E.
You have just constructed another parallelogram BECD with same base as ABCD and between the same parallels, as required.
In parallelogram ABCD with diagonal BD, we can prove that the two triangles ABC and CDB are congruent.
We also have sufficient evidence to prove that triangles CDB and BEC are congruent.
Now all three triangles are congruent to each other. They have equal areas.
Hello everyoneWhile this is true, it's a very special case. You should really consider two general parallelograms on the same base and between the same parallels, as I have indicated in the attached drawing. I outlined the proof in my first posting. Here are a few more details.
In the diagram, and are any two parallelograms on the base CD, with the points in a straight line.
Construct the diagonals and . Triangles and can be easily proven congruent; as can triangles and .
Drop perpendiculars from to meet at points . is then a rectangle. Therefore .
Area area area .
Grandad