Sunday, December 2, 2018

Geometry Problem 1404: Three Circles, Center, Intersecting Points, Collinearity, 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 Problem 1404: Three Circles, Center, Intersecting Points, Collinearity, Congruence, Tutoring.

Posted by Antonio Gutierrez at 3:53 PM
Labels: , , , , ,

4 comments:

  1. Sumith PeirisDecember 2, 2018 at 9:52 PM

    Let BA extended meet circle A at F

    Since BD = AC, ABCD is an isosceles Trapezoid with AB//CD.

    So < ACF = < AFC = < DCF = ϴ …… (1) say

    Hence < ACD = 2ϴ = < ABD and since ΔOBD ≡ ΔOBA (SSS), < DBO = < ABO = ϴ

    Now since < EBF = < ECF = ϴ = DCF from (1)

    It follows that C,D,E are collinear

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. c.t.e.oDecember 3, 2018 at 1:33 AM

    Extend DC to F ( F on B)
    Tr EAC = Tr BDF (congruent) => Tr EAD = Tr BCF => ang EDA = ang BCF

    ReplyDelete
  3. c.t.e.oDecember 3, 2018 at 9:50 AM

    BD = BA = AE = R and AD perpendicular to EB => DBAE rhombus
    => ang CDB + BDA + ADE = arc (CB)/2 + (BA)/2 + (AD + DC)/2 = 180

    ReplyDelete
  4. c.t.e.oDecember 10, 2018 at 9:35 AM

    From same point D (DBAE rhombus) we can draw just a paralelel to AB

    ReplyDelete

Subscribe to: Post Comments (Atom)