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.

2 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

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  2. 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.

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