Monday, March 23, 2020

Geometry Problem 1465: Tangential Quadrilateral, Incenter, Inscribed Circle, Equal Sum of Areas

Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Dynamic Geometry Problem 1465: Tangential Quadrilateral, Incenter, Inscribed Circle, Equal Sum of Areas, Step-by-step Illustration, iPad.

Posted by Antonio Gutierrez at 3:24 PM
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3 comments:

  1. c..t.e.oMarch 23, 2020 at 4:09 PM

    P1465
    AE=AH, BE=BF, CG=FC, DG=DH => AE+BE+CG+DG=AH+BF+FC+DH =>
    (AE+BE+CG+DG).R =(AH+BF+FC+DH).R
    S yellow = S green

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  2. Jacob HA (EMK2000)March 23, 2020 at 7:31 PM

    By tangent property, AH=AE, BE=BF, CF=CG, DG=DH
    Thus AD+BC=AB+CD

    On the other hand, the height of those triangles are the same, namely the radius,
    AD*r/2 + BC*r/2 = AB*r/2 + CD*r/2
    Hence, S1+S3 = S2+S4.

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  3. Sumith PeirisMarch 23, 2020 at 9:23 PM

    We have 4 pairs of congruent Tr.s, (APH,APE), (BPE,BPF), (CPF,CPG), (DPH,DPG) one each of which is yellow and the other green.

    The result is thus easily derived

    Sumith Peiris
    Moratuwa
    Sri Lanka

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