Marcia Bailey
Linda Crawford
Cynthia Dozier
ABC is an arbitrary triangle. On AB and BC arbitrary parallelograms ABDE and BCFG have been constructed. ED and FG have been extended to meet in H. Parallelogram ACKL has been constructed such that AL and CK are parallel to HB and have the same length as HB.

If follows that

area of ABDE + area of BCFG = area of ACKL.

Click here to see the GSP sketch. Verify that the above statement holds by dragging on the sketch.

Prove that the area of ACKL is equal to the sum of the areas of ABDE and BCFG.

PROOF:

Extend LA to intersect ED in I and extend KC to intersect GF in L. Extend HB to intersect AC in M and to intersect LK in N. Since HB is parallel to LA and KC, we now know IL, HN, and LC are parallel.

Then area of ABHI = area of ABDE since AB is the base of each and they have equal heights. Also area BCFG = area of BCLH since BC is the base of each and they have equal heights. Click here to see a visualization that these areas are equal.

Now AL = MN = CK = HB. So area of ALNM = area of ABHI since their bases NM and HB are equal and they have equal heights. Therefore, area of ABDE = area of ALNM.

Also area MNKC = area of BCLH since they have equal bases of HB and MN and also have equal heights. Thus, area BCFG = area of MNKC.

It then follows that

area ACKL = area of ALNM + area of MNKC = area of ABHI + area of ACLH.

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