Monday, July 16, 2012

Problem 784: Triangle, Orthocenter, Altitude, Midpoint of a side, Perpendicular

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 784.

Online Geometry Problem 784: Triangle, Orthocenter, Altitude, Midpoint of a side, Perpendicular.

3 comments:

  1. http://img402.imageshack.us/img402/4156/problem784.png

    Draw lines per attached sketch
    Note that P and Q are midpoints of BC’ and CB’
    Quadrilaterals MHEP and MFQH are cyclic
    In quadrilateral MHEP, ∠MEH=∠MPH=∠PHC’
    In quadrilateral MFQH , ∠MFH=∠MQH=∠QHB’
    ∆BHC’ similar to ∆ CHB’ => HC’/HB’=BC’/CB’=(.5.BC’)/(.5.CB’)=PC’/QB’
    So ∆PHC’ similar to ∆QHB’ …( case SAS)
    And ∠PHC’=∠QHB’
    ∠MEH=∠MFH => triangle MFE is isosceles => H is the midpoint of EF

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  2. Let O be the circumcenter of triangle ABC. Also, let M₁ be the midpoint of AH. From M₁ build E₁F₁ perpendicular to OM₁
    OMHM₁ is a Parallelogram
    OM₁ ǁ MH → E₁F₁ ǁ EF
    M₁→ midpoint [E₁F₁] →H midpoint of [EF]

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  3. consider problems 781,782,783 and the dilatation with center B and ratio 2.

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