Wednesday, August 29, 2018

Geometry Problem 1382: Circle, Diameter, Radius, Perpendicular, Congruence

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

Details: Click on the figure below.

Geometry Geometry Problem 1382: Circle, Diameter, Radius, Perpendicular, Congruence.
at

6 comments:

  1. AnonymousAugust 29, 2018 at 6:58 PM

    Angle COG = Angle FPE = 180 - Angle BOC
    E, O, F, P, are on the circle with center O1 and diameter OP
    Connect F with O1, and this line intersects the circle with center O1 at L.
    FL = OP = OC
    Angle FLE = Angle FPE
    triangle OCG = triangle FLE
    CG = EF

    Erina NJ

    ReplyDelete
  2. Peter TranAugust 30, 2018 at 9:00 AM

    https://photos.app.goo.gl/HnS668zUQ5v6RaBC9

    Note that quadr. OEPF is cyclic
    So ∠ (FOA)= ∠ (FPE)= u
    Draw PN ⊥ EF , N is the intersection of FN to circle OEPF
    Note that NE and OP are diameters of circle OEPF
    And ∠ (FNE)= ∠ (FPE)= u
    Triangle OCG congruent to ENF ( case HA) so CG= EF

    ReplyDelete
    Replies
    1. Erina NJAugust 31, 2018 at 8:45 AM

      Thank you for the picture of the geometry problem, and the comment you made to my solution.

      Delete
  3. AnonymousFebruary 12, 2022 at 5:50 AM

    Note that F,O,E,P concylic, so by Sine rule, EF = OPsin<P = OCsin<COG = CG. done
    Alex Toronto

    ReplyDelete

Share your solution or comment below! Your input is valuable and may be shared with the community.