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Geometry Proof - At it for Hours to No Avail

  1. Mar 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Given:

    Right triangle ABC with right angle A

    segment AD is an altitude of the triangle
    segment AE is an angle bisector of angle BAC
    segment AF is the median of side BC

    note: side BC has points D,E, and F on it in that order

    EDIT: I've added a picture of the figure:
    http://westchesterccb.com/proofpic.JPG [Broken]

    Prove:

    angle DAE is congruent to angle EAF



    2. Relevant equations

    N/A


    3. The attempt at a solution

    Got three similar triangles in ABD ~ DAC ~ DBA

    Sorry for the lack of formatting. To anyone who would attempt this proof and possibly solve it, I'd greatly appreciate it.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 30, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi meiso! Welcome to PF! :smile:

    Hint: that's the same as proving that angle BAD = angle FAC :wink:
     
  4. Mar 30, 2009 #3
    Yes, thank you for that hint. I had realized that and I still have not been able to solve it.
    I also have the equations (either from givens or similar triangles and substitution):

    angle ABC = angle DAE + angle EAF + angle FAC
    angle BAD + angle DAE = angle EAF + angle FAC = 45 degrees
    angle BAD = angle BCA

    If I could somehow prove angle ABC = angle BAD + angle DAE + angle EAF,
    then AF = BF, then BF = FC, and by the base angles theorem,
    angle FAC = angle FCA, then by the transitive property, angle BAD = angle FAC, and the rest is easy from there. However, I can't seem to establish that link above.
     
    Last edited: Mar 30, 2009
  5. Mar 30, 2009 #4

    tiny-tim

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    ah … well, that's why it's so important for you to start by telliing us what you have tried

    ok, new hint:

    you have a right-angled triangle, so draw it in a circle :wink:
     
  6. Mar 30, 2009 #5
    Wow. Very elegant tiny-tim. Thank you. BC becomes the diameter of the circle, so BF, FC, and AF are all radii, so their measures are all equal, which is what I needed.

    I was helping a high school student whom I tutor with this proof, and I never would have thought of using your method because he has not done anything like that in class. Thank you again, but if there is another method without using a circumscribed circle, I would love to know!
     
  7. Mar 30, 2009 #6

    tiny-tim

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    ok … alternative hint:

    draw FG parallel to AC :wink:
     
  8. Mar 30, 2009 #7
    Ok. So FG is drawn with G on segment BA, and intersects two sides of the triangle, so, being parallel to the base, it divides the other two sides proportionally.
    Since BF=FC, BF/FC = 1, so BG/GA must equal 1 and and therefore BG=GA.

    Following from that, I can use the SAS(BG=GA, two right angles, Reflexive GF) theorem to prove triangle BGF is congruent to triangle AGF. Triangle BGF is similar to triangle BAC (AA Theorem), so angle BFG = angle BCA. Angle AFG = angle FAC because they are alt. int. angles, but angle AFG also equals angle BCA (because angle BCA = angle BFG). By the transitive property (angle BCA = angle BAD already established), then, angle BAD = angle FAC!

    This all led from your hint. Please tell me if any of my logic was incorrect (I know it's a lot to sift through) or if there is a shorter way to reach the conclusion.

    And thanks again!
     
    Last edited: Mar 30, 2009
  9. Mar 31, 2009 #8

    tiny-tim

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    Yes that's ok, but from there on, because I know the circle method, I'd concentrate on sides rather than angles:

    FG perp BA, so (SAS) BF = AF, so CF = AF, so angle CAF = ACF :wink:
     
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