Sunday, October 8, 2017

Geometry Problem 1349: Three Squares, 90 Degrees, Perpendicular Lines, Diagonal. Math Infographic

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

Details: Click on the figure below.

Geometry Problem 1349: Three Squares, 90 Degrees, Perpendicular Lines, Diagonal. Math Infographic.

4 comments:

  1. Problem 1349
    Let < AHB = < HBG = < AGB = θ.

    AC2 = 2.CE2 = CE.CG, hence < CAE = < AGB = θ.

    Since < CBP = < CAP =θ,
    ABCP is concyclic and so
    < CPE = < ABC = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  2. Problem 1349
    In the extension of AB to B above point K such that AB=BK=>triangleKAD=triangleAFR=E.So
    KD=AE and KD perpendicular to AE.If the AE intersects the KD in Q and the BC in M. Then
    BM=MC and EQ/QA=EM/AD=3/2 =>EQ/EA=3/5 (1).Is EP/PA=BE/AH=2/3 =>EP/EA=2/5(2).
    From (1) and (2)=>EP/EQ=2/3=>EP/PQ=2/1=EC/CM=>CP//MQ=>CP is perpendicular in AE.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

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  3. https://photos.app.goo.gl/vydebtzBccTnATBf1

    Let BH meet EF at K
    Let AB=BC=CD=AD= 1
    So EK= 2/3 and AE= sqrt(5)
    Triangle ABP similar to EKP ( case AA)
    PE/PA= EK/AB= 2/3
    So PE=2/sqrt(5) and PA=3/sqrt(5)
    And PE/EB= 1/sqrt(5) and CE/AE= 1/sqrt(5)
    Triangle ECP similar to EBA ( case SAS)
    So ∠ (CPE)= ∠ (EBA)= 90 degrees

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