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CTK Exchange
sfwc
Member since Jun-19-03
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Jan-27-06, 05:56 PM (EST) |
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"Avoiding algebra."
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To see what is going on more easily, the algebra may be eliminated from the second proof at https://www.cut-the-knot.org/Curriculum/Geometry/SquaresAndTrianglesFromJapan.shtml in the following manner: Let S and T be chosen so that GFST is square, with the same centre as EKHB. So EFK, KGH, HTB, BSE, BCH and EAB are all congruent. Then DS is equal and parallel to TI so that DSIT is a parallelogram.
In particular, A(KDI) = A(KDSTI). Then: A(KMN) = A(KDI) (Vecten)
= A(KDSTI)
= A(EDK) + A(KESTH) + A(HKI)
= A(EAB) + A(KESTH) + A(HBC) (Vecten)
= A(BSE) + A(KESTH) + A(HTB)
= A(BEKH)
Note that it is not necessary to use G and H anywhere in this proof as S and T may be constructed independently. With a little work, you may be able to make this into a pww. Thankyou sfwc
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