Euclid's Elements Book 1: Proposition 35, Parallelogram Area

This is the thirty fifth proposition in Euclid's first book of The Elements. This proof shows that if you start with two parallelograms that share a base and end on the same parallel, they will be equal to each other (in area).

[@jakelm4256]

When I did this blind, my parallelograms overlapped on the top parallel. The proof was a little different, but it still worked.

[@EulersAcademy]

Thanks for watching.

[@carolineainsworth7618]

thanks bubba

[@EulersAcademy]

Thanks for watching. Glad I could help.

[@mnicole-l2n]

thanks bubba!! u rlly save lives ❤️

[@EulersAcademy]

Thanks for watching. Glad I could help.

[@Helixtronline]

Appreciated.

[@michaeljagdharry]

What if the second parallelogram was just a rectangle going directly up to AF from the base BC? This rectangle and the parallelogram ABCD would be unequal, as the respective interior angles between them would differ. Would anyone care to tell me what I am missing here?

[@OzoneTheLynx]

Probably too late for you, but for anyone else reading this. We only want to show that they have the same area. The proof is a little different. You get to the equality of the triangles ABE=DCF the same way. Now you don't have to remove and add those other triangles to get to the two paralleograms. Instead you'll find the triangles are just seperated by some trapezoid which you can add to the triangles. One triangle and the trapezoid gives you the parallelorgramm and the other triangle and the trapezoid gives you the rectangle.

[@Enlightenchannel]

Yea this confused me too. The language here is ambiguous. It should say that the area of the parallelogram is equivalent, not that the parallelogram is equivalent, because that gives you the impression that the interior angles are also equal, when they're obviously not. But I understand it now, thanks to this comment! LOL