AECP (Posted on 2011-03-01)

	Submitted by Bractals
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Lemma: Parallelograms ABCD and KLMN have the same area if 1) lines AB and KL coincide, 2) lines CD and MN coincide, and 3) |AB| = |KL|. Proof: [ABCD] = |AB|*[|BC|*sin(ABC)] = |AB|*[distance between sides AB and CD] = |KL|*[distance between sides KL and MN] = |KL|*[|MN|*sin(KLM)] = [KLMN] QED Construct ray PC intersecting lines AB and DE at points Q and R respectively. Construct ray EA intersecting line HI at point S and ray DB intersecting line FG at point T. Then [ABCD] = [AQRE] + [QBDR] = [QREA] + [RQBD] = [PCAS] + [CPTB] = [CASP] + [BCPT] = [CAHI] + [BCFG] QED Note: Use this theorem to prove the Pathagorean theorem.

	Subject	Author	Date
	Solution	Harry	2011-03-01 14:10:36