Monday, May 19, 2008

Elearn Geometry Problem 35: Quadrilateral, Circle

Incenters and Inradii in Cyclic Quadrilateral

See complete Problem 35 at:
www.gogeometry.com/problem/p035_incenter_cyclic_quadrilateral.htm

Incenters and Inradii in Cyclic Quadrilateral. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 9:05 PM
Labels: , , , ,

5 comments:

  1. 배덕락(bae deok rak)October 6, 2010 at 9:17 PM

    (1) Since E is an incenter of triangle BD,
    ang(AED)=90°+(ang(ABD)/2)=90°+(ang(ACD)/2),
    ang(HED)=ang(HAD)=ang(CAD)/2,
    ang(BEA)=90°+(ang(BDA)/2)=90°+(ang(BCA)/2) and
    ang(FEB)=ang(FAB)=ang(CAB)/2.
    Hence we see that
    ang(FEH)
    =360°-(ang(AED)+ang(HED)+ang(BEA)+ang(FEB)
    =360°-
    (180°+(ang(ABD)+ang(CAD)+ang(BDA)+ang(CAB))/2
    =360°-(180°+(ang(ABC)+ang(ADC))/2
    =360°-(180°+90°)
    =90°
    and hence EFGH is a rectangle.

    (2) We get
    (i) By (1), EH^2 =FG^2.
    (ii) EH^2
    =(r_d -r_a)^2 +(AD-(AB+AD-BD)/2-(DA+DC-AC)/2 )^2
    (iii) FG^2
    =(r_c -r_b)^2 +(BC-(BA+BC-AC)/2-(CB+CD-BD)/2 )^2

    Hence we have (r_d -r_a)^2 =(r_c -r_b)^2,
    that is, r_a +r_c =r_b +r_d

    ReplyDelete
  2. Nilton LapaMarch 3, 2012 at 2:54 AM

    I'd like to know why is it ang(HED)=ang(HAD), in the solution presented by bae deok rak. I thank you in advance.

    ReplyDelete
  3. AnonymousNovember 4, 2018 at 3:12 AM

    ang(DBA)=ang(DCA) hence ang(DHA)=ang(DEA) therefore AEHD concyclic hence ang(HED)=ang(EAD)

    ReplyDelete
  4. GregDecember 11, 2018 at 10:40 AM

    Correction to my earlier comment of Nov 4, 2018 : read HAD and not EAD at the end

    ang(DBA)=ang(DCA) hence ang(DHA)=ang(DEA) therefore AEHD concyclic hence ang(HED)=ang(HAD)

    ReplyDelete
  5. Stan FULGERDecember 21, 2022 at 4:32 AM

    My solution, 100% synthetic, at my blog, https://stanfulger.blogspot.com/2022/12/problem-35-gogeometry.html

    ReplyDelete

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