Sunday, January 25, 2009

Elearn Geometry Problem 233: Parallelogram, Exterior line, Perpendicular lines

Problem: Parallelogram, Exterior line, Perpendicular lines

See complete Problem 232 at:
gogeometry.com/problem/p233_parallelogram_perpendicular_lines.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

6 comments:

  1. Let be O the intersection of AC and BD, and O' the projection of O on the exterior line. Let denote by o the lenght of OO'.
    In the trapezoid BB'D'D :
    o=(b+d)/2
    In the trapezoid AA'C'C :
    o=(a+c)/2.
    From this relations we have that :
    a+c=b+d

    ReplyDelete
  2. like P234, 235

    draw AB", DC" // A'C'

    b-a /AB = c-d/CD
    b-a = c-d => b+d = c+a

    ReplyDelete
  3. Draw D'D''DD''' parallel to A'B'D'C' thro' D.
    where D',D'',D''' lie on AA', BB', CC' resp.
    now, D'A' = D''B' = DD' = D'''C'.
    so, it requires us to prove that AD' + CD''' = BD'' which comes from problem 232.

    ReplyDelete
  4. Proof follows easily from my proof of Problem 232

    ReplyDelete
    Replies
    1. Extend A`A to form a rectangle with C (A'ECC'). Let BB' meet EC at F => BF=b-c
      Join AC to form another rectangle A'ACC'. Let DD' meet AC at G => DG=a-d
      BFC and DGA are congruent => BF=DG and hence a+c=b+d

      Delete
  5. Draw a line at A"B"C" pass through D // A'B'D'C'
    From problem 232, AA"+CC"=BB" & d=A"A=B"B=C"C
    a+c=AA"+A"A'+CC"+C"C
    =AA"+CC"+A"A"+C"C'
    =BB"+d+d
    =BB"+B"B'+d
    =b+d

    ReplyDelete