Re: Geometry problem...

From: Philip Holman (namlohip_at_comcast.net)
Date: 11/26/04 

Date: Thu, 25 Nov 2004 21:42:38 -0800

"Philippe 92" <nospam@free.invalid> wrote in message
news:mn.d0047d4bd61a7b3e.22155@free.invalid...
> 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".

That's a nice symmetrical quadrilateral you drew. Does it come down to
perpendicular diagonals always result in symmetrical quadrilaterals?

PH

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