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two triangle construction problems (thanks)

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  • Luís Lopes
    Dear Nikos, Thank you very much for this! The solution is exactly what I was looking for!!! Best regards, Luis ...
    Message 1 of 1 , Aug 5, 2002
      Dear Nikos,

      Thank you very much for this! The
      solution is exactly what I was looking for!!!

      Best regards,
      Luis

      >From: "Nikolaos Dergiades" <ndergiades@...>
      >Reply-To: Hyacinthos@yahoogroups.com
      >To: <Hyacinthos@yahoogroups.com>
      >Subject: Re: [EMHL] two triangle construction problems
      >Date: Sat, 27 Jul 2002 10:41:20 +0300
      >
      >Dear Luis,
      >
      > >Secondly, I have been trying to solve the
      > >following two problems for a while without
      > >success:
      > >
      > >i) A, a-b, a-c
      > >ii) A, a-b, b-c
      > >
      > >I know of A, a+b, a+c from Court's book but
      > >even so couldn't mimic a similar solution
      > >(the kind I am looking for) to i).
      >
      >
      >********
      >Since a-c = (a-b) + (b-c) problem ii) leads to problem i)
      >
      >Problem i) has solution analogous to problem (A, a+b, a+c).
      >More precisely taking on BA, BB' = a and on CA, CC' = a
      >has known A, AB' = a-c, AC' = a-b and can be constructed.
      >Now on the sides AB', AC' of a known triangle we want to
      >construct the points B, C such that B'B = BC = CC' = x.
      >
      >This construction results from the following:
      >If vector BB'' = vector CC' and the line B'B'' meets the
      >line
      >AC at D and the parallel from D to BC meets B'C' at E then
      >the parallelogram BB''C'C is rhombus BB'' = B''C' = x and
      >B''C'/DE = B'B''/B'D = BB''/AD = BB'/AB' or
      >x / DE = x / AD = x / AB'. Hence AD = DE = AB'.
      >
      >Construction: On AC' we take D such that AD = AB'.
      >The circle with center D and radius AD meets B'C' at E
      >(there are perhaps two, we select the appropriate).
      >The parallel from C' to DE meets B'D at B'', the parallel
      >from B'' to AC' meets AB' at B and the parallel from B to
      >B''C' meets AC' at C.
      >
      >Best regards
      >Nikos Dergiades
      >


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