Saturday, February 19, 2011

Problem 580: Triangles, Equal Angle, Transversal, Product of Sides, Areas

Geometry Problem
Click the figure below to see the complete problem 580.

Problem 580: Triangles, Equal Angle, Transversal, Product of Sides, Areas.
Zoom at: Geometry Problem 580

Posted by Antonio Gutierrez at 3:51 PM
Labels: , , ,

4 comments:

  1. c .t . e. oFebruary 20, 2011 at 12:33 AM

    draw DH1, AH altitudes on BE, BC
    BE∙ DH1 = S1, BC∙AH = S => S1/S = (BE∙DH1)/(BC∙AH)
    from ▲DBH1, ▲BAH => DH1/AH = BD/AB substitute above
    S1/S = (BE∙BD)/(BC∙AB)

    (I have sent yesterday solution of 579)

    ReplyDelete
  2. AnonymousFebruary 20, 2011 at 1:09 AM

    Join point E & point A. Then area of triangle BED (S1) is proportional to area of triangle BEA (S2), and the ratio is:
    S1/S2 = BD/BA. (1)
    Now area of triangle BAC (S) is proportional to area of triangle BEA (S2): S/S2 = BE/BC. (2)
    We multiply (1) & (2) to obtain the result:
    (S1/S2)·(S/S2) = S1/S = BD·BA/(BE·BC).

    MIGUE.

    ReplyDelete
  3. AnonymousFebruary 20, 2011 at 7:39 AM

    Sorry:
    S/S2=BC/BE
    (S1/S2)•(S2/S)=BD•BE/(BA•BC)

    ReplyDelete
  4. AnonymousFebruary 22, 2011 at 1:43 AM

    S=1/2.BA.BC.sinB
    S1=1/2.BD.BE.sinB

    ReplyDelete

Subscribe to: Post Comments (Atom)