Re: Geometry problem...

From: Philippe 92 (nospam_at_free.invalid)
Date: 11/25/04 

Date: Fri, 26 Nov 2004 00:04:31 +0100

Hello,

Philip Holman wrote :
> "Michael Swift" <miswift2000@yahoo.com> wrote in message
> news:c5508f88.0411250749.1a0ea5b5@posting.google.com...
>> Hi,
>>
>> I can't solve the following problem. I'd be grateful for hints
>>
>> Circle with center in point H is inscribed into convex quadrilateral
>> ABCD, point H doesn't lie on line AC. Diagonals AC and BD intersect at
>> point F. Line passing through point F and perpendicular to line BD,
>> cuts lines AH and CH in points R and S respectively. Prove that RF=FS.
>>
>
> It doesn't. AC and BD can perpendicular to each other in which case RF=AF and
> FS=FC where AF is not equal to FC.
>

False.
AF must be = FC for ABCD to be tangential, H not on AC, and AC _|_ BD.

Let a triangle PAB have an incircle with incenter H and let the
incircle be tangent to
PA at U, and PB at V. Then the lines BH, UV, and the perpendicular to
BH through
A concur in a point F (Honsberger 1995).
http://mathworld.wolfram.com/Incircle.html
[points renamed to original post's names]

http://chephip.free.fr/img/figholman.gif
Let a secant CD, C on PA, D on PB with CD tangent to incircle, to get a
tangential quadrilateral ABCD, tangent to ABCD in T, U, V, W.
T on CD, U on BC, V on AD W on AB.

>From the Brianchon theorem, it's easy to proove that the four lines
AC, BD, UV, TW concur in F.

Let's choose CD such as the diagonals BD _|_ AC.
FB _|_ FA, hence the locus of F is the circle with diameter AB.
This circle intersects UV at F and F', with FB going through H and F'A
going
through H from the above property.
F' is rejected as "point H doesn't lie on line AC".

Hence BD is angle bissector of angle B (and D) and FC = FA.

BTW it's easy to get that FA=FC=FR=FS is here *not* equal to the
incircle radius,
to counteract Narasimham's "hint".

But just now, I don't have a proof of Honsberger property, and I don't
have any
Idea in the general case (BD and AC not perpendicular, R!=A, S!=C).

Regards. 

-- 
philippe
(chephip at free dot fr)

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