13436Re: Concurrencies on a triangle.
- Jun 27, 2006--- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...>
> ...Dear all my friends.
>7f. As a helping proposition, we use the result that the points B',
>A, S, are collinear ( we shall prove the proposition for this
>result, below ). Hence, we consider the segments ...
A friend of mine Kostas Helatiotis, said me that we don't need
any helping proposition for the proof of (7), in message 13274.
7f. From cyclic non-convex hexagon D'D''D1CEB, by Pascal's
theorem, we have that the points B', A, S, are collinear. Hence, we
consider the segments ...
I apologize for my blindness and please, cancel all about the
proof of the helping proposition (7.1), as unnecessary prattlings.
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