two triangle construction problems (thanks)
- Dear Nikos,
Thank you very much for this! The
solution is exactly what I was looking for!!!
>From: "Nikolaos Dergiades" <ndergiades@...>_________________________________________________________________
>Subject: Re: [EMHL] two triangle construction problems
>Date: Sat, 27 Jul 2002 10:41:20 +0300
> >Secondly, I have been trying to solve the
> >following two problems for a while without
> >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
>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.
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