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.

1 comment:

  1. 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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