Saturday, September 5, 2009

Problem 352. Tangential quadrilateral, Incircles, Common tangent, Circumscribable or Tangential quadrilateral

Proposed Problem
Click the figure below to see the complete problem 352 about Tangential quadrilateral, Incircles, Common tangent, Circumscribable or Tangential quadrilateral.

Problem 352. Tangential quadrilateral, Incircles, Common tangent, Circumscribable or Tangential quadrilateral.
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Complete Problem 352
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 12:45 PM
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3 comments:

  1. c .t . e. oJanuary 26, 2010 at 11:35 AM

    name S circle 1 meet AB, P meet AE, H meet BC
    name J circle 2 meet BC, Y meet ED, V meet DC

    ABCD is tangential quadrilateral so =>

    AD + BC = AB + DC
    AD+MN+FM+NG+BH+JC = AS+BS+DV+CV (HJ = FG, see P355)
    AD+MN+FM+NG = AS+DV (BH=BS, JC=CV tang from a point)

    AD+MN+FM+NG = AM+MP+DN+NT ( AS=AP=AM+MP,DV=DT=DN+NT)

    AD + MN = AM + DN (FM=MP,NG=NT)

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  2. c .t . e. oJanuary 26, 2010 at 2:15 PM

    second row have to read
    ... T meet ED, ...

    ReplyDelete
  3. Peter TranDecember 25, 2015 at 9:11 PM

    http://s3.postimg.org/4tcvl07pv/pro_352.png

    ABCD is tangential quadrilateral (see skech)
    So AB+CD=BC+AD … (1)
    AM=AP-u =AB-x-u
    DN=AS-v=CD-y-v
    So AM+DN=AB+CD-(x+y)-(u+v)
    MN=FG-u-v=QR-u-v=BC-(x+y) -(u+v)
    So MN+AD=BC+AD-(x+y)-(u+v)
    Replace BC+AD=AB+CD we get AM+DN+MN+AD => quadrilateral AMND is tangential

    ReplyDelete

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