A perspector question
- Dear Hauke, I think the answer is yes.
If we consider an arbitrary convex quadrilateral ABCD, we can construct
a point P inwardly to it, such that the triangles PAB, PDC to be
similar, with for example <PAB = <PCD and <PBA = <PDC, as well known.
( The point P is the intersection point of two Appolonian circles with
respect to the diagonals AC, BD of ABCD and ratio r = AB / CD ).
Let now S be, the intersection point of the sidelines AD, BC and we
define three points P', C', D', on the rays SP, SC, SD respectively,
such that P'C' // PC and P'D' // PD.
The triangles PAB, P'D'C', are perspective and similar as the problem
Best regards, Kostas Vittas.
--- In Hyacinthos@yahoogroups.com, "Hauke Reddmann" <fc3a501@...> wrote:
> Let ABC and A'B'C' be perspective trangles.
> Can ABC and A'B'C' be similar *without*
> AB||A'B' etc.?
- --- In Hyacinthos@yahoogroups.com, "Kostas Vittas" <vittasko@...> wrote:
>A bit of computing also affirms that -
> Dear Hauke, I think the answer is yes.
a triangle similar to ABC can be rotated
(1), scaled (1) and shifted (2 free variables),
and the perspector just demands one equation.
By ray law (or what it's called in English)
the line through the barycenters M and M' of
ABC and A'B'C' also goes through O. In the
"standard" case of course a line through *any*
two centers will go through O but it still
would be interesting if, say, H,H' and O are
collinear (H orthocenter) and still the sides
of ABC and A'B'C' not parallel. Remember, there
are still 3 variables to use up :-)