Wednesday, July 8, 2015

Geometry Problem 1130: Cyclic Quadrilateral, Diagonals, Angle Bisector, Congruent Angles

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1130: Cyclic Quadrilateral, Diagonals, Angle Bisector, Congruent Angles.

Posted by Antonio Gutierrez at 9:21 PM
Labels: , , ,

6 comments:

  1. Peter TranJuly 9, 2015 at 11:36 AM

    http://s4.postimg.org/9k94xkj65/pro_1130.png

    Let AD cut BC at N and EF cut BC and AD at L and M
    From N make line NG perpendicular to EF
    Observe that M, N are harmonic conjugate points of A and D and
    L, N are harmonic conjugate points of B, C
    So circumcircles of triangles MGN and LGN are Apollonius circles of AD and BC
    And GM and GL are internal bisectors of angles AGD and BGC
    So ∠ (AGB)= ∠ (DGC)

    ReplyDelete
  2. Peter TranJuly 9, 2015 at 5:37 PM

    Per my solution as above , qudrilateral ABCD may not need to be cyclic to get the result

    Peter Tran

    ReplyDelete
    Replies
    1. Antonio GutierrezJuly 10, 2015 at 2:50 PM

      Problem 1130 to Peter: in your 2nd statement, why NG is perpendicular to EF? Thanks.

      Delete
    2. Peter TranJuly 11, 2015 at 9:40 PM

      To Antonio
      Please see my response below.

      1. L and N are harmonic conjugate with respect to B and C
      Let circle diameter LN (Apollonius circle of B and C) cut EF at G’.
      Since G’ is located on Apollonius circle of B and C so G’EF is an angle bisector of angle BG’C and G’EF ⊥ NG’
      2. N and M are harmonic conjugate with respect to A and D
      Since NG’⊥ G’M => G’ locate on Apollonius circle of A and D ( diameter MN)
      And G’M is an angle bisector of ∠AG’D =>angle ∠AG’B congruent to ∠CG’D.

      G’ is an only point of line EF have this property since G’ is the perpendicular projection of N over EF. G’ in fact coincide to G .

      Conclusion: for a general quadrilateral ABCD, we will have one and only one point G on the EF so that EF is the angle bisector of angles BGC and AGD. At the point G , we will have angle AGB congruent to CGD.
      See link below for the typical non-cyclic quadrilateral ABCD and location of point G satisfied the problem statement

      http://s12.postimg.org/9cy9rp26l/pro_1130test.png

      Delete
  3. AmoreiraSeptember 1, 2015 at 7:21 AM

    This comment has been removed by the author.

    ReplyDelete
  4. Ivan BazarovDecember 20, 2015 at 2:26 PM

    Let circumcircles of ABE and DCE meet at G'. Radical axis of these circles is G'E. Because FB*FA=FC*FD, F is on line G'E. Note that <BAE=<BG'E and <CDE=<CG'E, but since <BAE=<CDE, <BG'E=<CG'E. G' lies on EF so G and G' concur. With this fact, <AGB=<AEB=<CED=<CGD.

    ReplyDelete

Subscribe to: Post Comments (Atom)