Monday, May 19, 2008

Elearn Geometry Problem 37



See complete Problem 37
Right triangle, altitude, incenters, orthocenter. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

4 comments:

  1. join A → F → E, C → G → E
    extend FE → E' (on BG), GE → E" (on FB)

    ang ABD = C, ang DBC = A ( perpendicular sides )

    ▲CE"G =>
    ang E"CB + ang E"BC = C/2 + A + C/2 = C + A = 90°

    => CE"B = 90°
    -----------------------------------------
    ▲'AE'B =>
    ang E'AB + ang E'BA = A/2 + C + A/2 = A + C = 90°

    => AE'B = 90°
    -----------------------------------------

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  2. In problem 25, BE and FG are congruent and perpendicular.
    Let FP be an altitude on the base BG.
    Then we get
    (1) ang(A) +ang(C)=90 degree
    (2) ang(GFP)=ang(A)/2 +ang(C)/2=45 deg
    (3) ang(FGP)=ang(C)/2 + (90- ang(C))/2=45 deg.
    Hence we see that ang(EPG)=90 degree and so
    E is an orthocenter of triangle BFG.

    ReplyDelete
  3. Extend BF &BG to M and N on AC ( See sketch)
    Note that AE, CE , BF and BG are angle bisectors , A,F,E and E,G, C are collinear
    ∠ (M)= ∠ (MBC) …. Both angles complement to ∠ (MBD)
    ∆ MBC is isosceles => angle bisector CE ⊥ to the base BM
    Similarly AE ⊥ to BM => E is orthocenter of ∆BFG

    ReplyDelete
  4. See below for the sketch
    http://img98.imageshack.us/img98/9435/problem37.png

    ReplyDelete