Saturday, November 14, 2015

Geometry Problem 1166: Triangle, Circles, Diameters, Secant, Midpoint, Congruence

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

Click the figure below to view more details of problem 1166.

Online Math: Geometry Problem 1166: Triangle, Circles, Diameters, Secant, Midpoint, Congruence.

Posted by Antonio Gutierrez at 12:51 PM
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4 comments:

  1. Stan FulgerNovember 15, 2015 at 2:25 AM

    Take X - reflection of E in F, AECX is a parallelogram, so AX||CE, i.e. AX_|_BE, A-X-D collinear, XED is right-angled triangle, thus EF=FX=DF, done.

    Best regards

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    Replies
    1. Sumith PeirisNovember 15, 2015 at 9:06 AM

      Excellent work Stan!

      With your permission a slightly modified approach...,,,,,

      Let EF meet AD at X then Tr.s AXF & CEX are congruent ASA

      So XF = FE and so F is the centre of right Tr. XDE and hence DF = FE

      Delete
  2. PravinNovember 17, 2015 at 8:46 AM

    Corollary: Let BF(extended) meet the circles at X,Y.Then XY is bisected at F.

    ReplyDelete
  3. Ivan BazarovNovember 25, 2015 at 10:18 AM

    Call other intersection of the 2 circles B' which is the foot of altitude from B.
    Let EF and AD meet at P. Then <PAF=<DBB'=<ECB', which means that AP||EC.
    From congruent triangles APF and CEF, PF=FE, which makes F circumcenter of right triangle PDE.

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