Monday, July 6, 2015

Geometry Problem 1128: Right Triangles, Medians, Parallel, Angles, Congruence, Half the measure

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1128: Right Triangles, Medians, Parallel, Angles, Congruence, Half the measure.

Posted by Antonio Gutierrez at 5:05 PM
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5 comments:

  1. Peter TranJuly 6, 2015 at 9:02 PM

    http://s7.postimg.org/ccpzrcdez/pro_1128.png

    Observe that quadrilateral DBEB1 is cyclic so α = α 1
    Let B1M1 cut BE extension at N
    Let b=∠ (MBC) => ∠ (MCB)= β
    Since BM//B1M1 => ∠ (M1Nx)= β
    And ∠ (CFE)= 180- β - α
    And ∠ (M1B1C1)= β - α =∠ (M1C1B1)
    In triangle GFC1 θ = 180-(180- β - α)-( β - α)= 2. α

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  2. Sumith PeirisSeptember 10, 2019 at 10:59 PM

    Let BE extended meet M1B1 at Y and A1B1 extended at X

    alpha1 = alpha since BDB1E is concyclic

    If < BAC = p, then < BXD = 90-alpha
    So < B1YX = 180-2.alpha-2p = < MBC = 90- p since BM//BM1

    Hence 2.alpha + p = 90

    Now consider quadrilateral AGA1D

    180-p + 180- theta + 2.alpha + p = 360

    Therefore Theta = 2.alpha

    Sumith Peitis
    Moratuwa
    Sri Lanka

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  3. Sailendra TApril 22, 2021 at 12:59 PM

    For ease of understanding, read @ as Alpha
    Extend AB to meet C1F at H
    Observe that HB1D and HBE are similar => m(BEH)=m(BDB1)=@ -------------(1)
    Let m(BAC)=A and extend BE to meet M1B1 at P
    m(M1B1E)=180-m(MBE)=90+A=>m(M1C1B1)=m(M1B1C1)=m(EB1P)=90-A-@
    Extend C1G to meet AB at Q => m(AQG)=m(M1C1B1)+m(AHC1)=90-A-@+90-@=180-A-2@
    Therefore in Tr.AGQ, m(AGQ)=2@=m(CGC1) -----------------(2)

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