Wednesday, July 15, 2015

Geometry Problem 1136: Triangle, Circumcircle, Orthocenter, Midpoint, Arc, 90 Degrees, Angle

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 1136: Triangle, Circumcircle, Orthocenter, Midpoint, Arc, 90 Degrees, Angle.

2 comments:

  1. http://s17.postimg.org/3wo281qsv/pro_1136.png

    Draw diameter BE ; connect AE and EC
    EC⊥BC => EC//AH
    EA⊥BA => EA//CH
    So AHCE is a parallelogram => H, M, E are collinear
    So BD⊥ MD

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  2. Extend HM to X such that HM = MX, so AHCX is a parallelogram

    Hence < AHC = A + C = < AXC, hence X must lie on circle ABC {B + (A+C) = 180}

    Therefore BX is a a diameter passing thro O and so < BDH = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

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