Friday, December 25, 2009

Problem 408: Cyclic quadrilateral, Perpendicular, Parallelogram, Congruence

Proposed Problem
Click the figure below to see the complete problem 408 about Cyclic quadrilateral, Perpendicular, Parallelogram, Congruence.

Problem 408: Cyclic quadrilateral, Perpendicular, Parallelogram, Congruence.
See also:
Complete Problem 408
Level: High School, SAT Prep, College geometry

5 comments:

  1. extend BF to P, and CG to Q ( P and Q on circle )

    a) ang NBC = ang FPQ ( BP//CQ, and BPQC cyclic) (1)

    b) join A to Q, ang ADE = ang AQG ( have same arc ABC)
    => ang EAG = ang GAQ
    => GQ = GM (2)

    c) at the same way
    => NF = FP (3)

    d) from (2) and (3)
    ang MNF = ang FPQ (4)

    e) from (1) and (4)
    ang NBC = ang FNM
    => BC//NM

    ReplyDelete
  2. can you tell me, please
    why my solution is deleted

    ReplyDelete
  3. Accidentally deleted.

    Following the solution posted by c.t.e.o on December 27, 2009.

    extend BF to P, and CG to Q ( P and Q on circle )

    a) ang NBC = ang FPQ ( BP//CQ, and BPQC cyclic) (1)

    b) join A to Q, ang ADE = ang AQG ( have same arc ABC)
    => ang EAG = ang GAQ
    => GQ = GM (2)

    c) at the same way
    => NF = FP (3)

    d) from (2) and (3)
    ang MNF = ang FPQ (4)

    e) from (1) and (4)
    ang NBC = ang FNM
    => BC//NM

    ReplyDelete
  4. thanks

    the main idea of solution was

    finding AD as simetrical line
    =>
    everything under it, is congruent to everything above it
    ( segment, angle, triangle, ... )

    ReplyDelete
  5. Problem 408
    The points M, N are orthocenter of triangles ACD, ABD respectively. Bring OP perpendicular AD.Then BN=//2OP and CM=//2OP. So BN=//CM, therefore BCMN is parallelogram.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete