Sunday, August 21, 2016

Geometry Problem 1251: Triangle, Circle, Diameter, Perpendicular, 90 Degrees

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1251.


Geometry Problem 1251: Triangle, Circle, Diameter, Perpendicular, 90 Degrees

2 comments:

  1. https://goo.gl/photos/4guoy5kN3aqGP59PA

    Draw circle O1 diameter GB and circle O2 diameter HB
    Note that circle O1 pass through points F and D
    Circle O2 pass through points F and E
    We have ∠( GBF)= ∠(GDF)= x
    ∠(FBH)= ∠(FEH)=y
    But ∠(FEC)= ∠(FDC)=y
    But ∠(GDF)+ ∠(FDC)=90= x+y
    So ∠ (GBH)= ∠(GBF)+ ∠(FBH)=x+y=90

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  2. Note that GBDF, AEDF and BEFH are all cyclic quadrilaterals.

    Hence < GBF = < GDF = < AEF = < BHF

    So < GBH = < GBF + < HBF = < BHF + <HBF = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

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