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Triangle question

  1. Jun 29, 2009 #1
    1. The problem statement, all variables and given/known data
    let [tex]\Delta[/tex]ABC be a right angled triangle such that angle A = 90o, AB = AC and let M be the mid point of the side AC. Take the point P on the side BC so that AP is vertical to BM. Let H be the intersection point of AP and BM. Find the ratio BP : PC


    2. Relevant equations



    3. The attempt at a solution
    Please give me a clue to start
     
  2. jcsd
  3. Jun 29, 2009 #2

    dx

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    Do you mean perpendicular?
     
  4. Jun 29, 2009 #3
    Draw a picture
     
  5. Jun 30, 2009 #4
    yes it's perpendicular

    i have drawn a picture

    because CM = MA, is it correct if i assume that angle CMB = angle AMB = 22.5o?

    thx
     

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  6. Jun 30, 2009 #5
    If AB=AC then you got isosceles triangle and the angles are: 90,45,45 degrees. Also P divides BC on two equal parts (since the triangle is isosceles).

    Using sine and the Pitagorean theorem you would easly come up with AH and AM. Since AM=AC/2, you will find AC.

    Regards.
     
    Last edited: Jun 30, 2009
  7. Jul 1, 2009 #6
    sorry but the question is BP : PC

    if P divides BC on two equal parts, it means that BP : PC = 1 : 1
    but the answer is 2 : 1 and i don't know how to get it

    is angle CMB = angle AMB = 22.5 degree?

    thx
     
  8. Jul 1, 2009 #7
    1. A(0,0); B(a,0); C(0,a), M(0,a/2).
    2. Slope of MB = -1/2
    3. Slope of AP = 2
    4. Equation of AP: y = 2x
    5. Equation of CB : y = -x + a
    6.Intersection of AP & CB is P(a/3, 2a/3).
    7. BP: PC = 2:1.
     
  9. Jul 2, 2009 #8
    wow, i never thought of using line equation and gradient
    really nice...

    thx a lot leong ^^
     
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