This a better problem than that FE: Geometry

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This a better problem than that FE: Geometry

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#1 h Sep 2, 2016, 9:51 PM • 2 Y

Y by Adventure10, Mango247

Let ABCD be a quadrilateral. If a circle can be inscribed in it, prove that AB +CD= BC+DA. (The Pitot theorem)

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#2 h Sep 2, 2016, 9:52 PM • 2 Y

Y by Adventure10, Mango247

You put in the easy version. The better statement is quadrilateral $ABCD$ is circumscriptable if and only if $AB+CD=AD+BC.$

First, we prove the "if". Let the incircle meet $AB, BC, CD, DA$ at $E,F,G,H$ respectively. Then, $AE=AH, BE=BF, CF=CG, DG=DH.$ Summing then combining gives $AB+CD=AD+BC.$

Now, we assume $AB+CD=AD+BC$ and prove $ABCD$ is crcumscriptable. Construct circle $O$ tangent to $AD, DC, BC.$ We generate point $B'$ on $BC$ such that $AB'$ is tangent to circle $O$ as well. Thus, by the "if" statement, $AB'+CD=AD+B'C.$ By hypothesis, $AB+CD=AD+BC.$ Thus, $AB-AB'=BC-B'C.$ Note that since $BC-B'C=BB',$ we have $AB-AB'=BB'.$ By the triangle inequality, this means $B'$ lies on $AB.$ However, $B'$ also lies on $BC,$ so $B'=B$ as we wished.

This post has been edited 1 time. Last edited by checkmatetang, Sep 2, 2016, 9:53 PM
Reason: more info!

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#3 h Sep 2, 2016, 9:56 PM • 2 Y

Y by Adventure10, Mango247

What does circumscriptable mean? Do you mean circumscribable? I knew, how to prove it, it was mainly redemption for my ridiculous problem I posted earlier.

This post has been edited 1 time. Last edited by First, Sep 2, 2016, 9:58 PM

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#4 h Sep 2, 2016, 9:58 PM • 2 Y

Y by Adventure10, Mango247

No, circumscriptable is the same as tangential, which means a circle can be inscribed in the figure.

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tree3

#5 h Sep 3, 2016, 1:05 AM • 2 Y

Y by Adventure10, Mango247

Consider the intersections of the quadrilateral and circle. For distinct intersections each by two tangents implies that there are four sets of tangent lines. Let them be w,x,y, and z. Now w and x, say are on one side, x and y on another, y and z on another, and z and w on the last. Now w+x+y+z, summing up two opposite sides equals itself. Q.E.D.

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jayme

#6 h Sep 3, 2016, 6:54 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,

see
http://jl.ayme.pagesperso-orange.fr/Docs/Urquhart.pdf p. 6...
and also

http://jl.ayme.pagesperso-orange.fr/Docs/Demir.pdf p. 7-11.

Sincerely
Jean-Louis

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#7 h Sep 3, 2016, 1:30 PM • 1 Y

Y by Adventure10

I just used a bunch of congruent tangents, added up their measures and matched up numbers.

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