Saturday, June 28, 2008

Elearn Geometry Problem 132



See complete Problem 132
Triangle, Angle,60, Orthocenter, Congruence, Midpoint. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

2 comments:

  1. To prove the proposed theorem, two steps are needed.

    Firstly, prove that EH is half of BD (E is the midpoint of BD).

    Secondly, prove that the triangle AHE is equilateral. It follows that AH = EH = (1/2)BD.

    The details are as follows:

    Connect A and D, D and H.
    Extend CA to F.

    Then ADHC is a parallelogram(as M is the midpoint of both line segments of AH and CD).

    mDAB is a right angle(as mDAF = mHCA = mABH = 30 degrees, and mBAC = 60 degrees).

    mDHB = 90 degrees (as DH is perpendicular to AC).

    In right triangle DAB, AE = (1/2)BD.

    In right triangle DHB, EH = (1/2)BD.

    So triangle AEH is an isosceles triangle.

    Also note that quadrilateral DAHB is cyclic (as mDAB and mDHB are right angles), we have:
    mEAB = mEBA = mDHA = mHAC.

    So mEAH = mEAB + mBAH = mHAC + mBAH = mBAC = 60 degrees.

    So isosceles triangle EAH is equilateral.

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  2. Let CH cut BA at X, BH cut AC at Y, BA cut DH at Z

    ACHD is a //gram, so mDAB=mAXC=mBYA=mBHD
    A,D,B,H concyclic and ZHA similar to ZBD
    HA/BD=ZH/ZB=AY/AB=1/2

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