Friday, November 6, 2015

Geometry Problem 1162: Triangle, Congruence, Double Angle, Parallel Lines, Circle, Circumcenter

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
Click the diagram below for more details.

Online Math: Geometry Problem 1162: Triangle, Congruence, Double Angle, Parallel Lines, Circle, Circumcenter.

3 comments:

  1. Let a=18
    Because triangles ABE and AFC are both isosceles, BF perpendicularly bisects AE, meaning that
    <ABF=<ABE/2=4a/2=2a=36
    Because <BAF=4a=72, and <BFA=180-<ABF-<BAF=180-36-72=72, BA=BF=BE or that B is circumcenter of AFE.
    We can now say that <FBC=2a=<BCF so BF=CF. Combined with BD=EC, triangles FBD and FCE are congruent and DF=FE. Thus <DEF=<EDF=4a and <DFE=2a.
    Finally, <DFC=<DFE+<EFC=2a+2a=4a=<BAC so BA //DF

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  2. On what basis is a = 18? This needs to be proved

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  3. Sorry, my full proof must have been cut somehow
    Let <DAE=<EAC=a.
    Because AF=FE, <EAF=<AEF=a and <EFC=2a
    Because FE=EC, <ECF=<ECF=2a
    Because <DAC=<DAE+<EAC=a+a=2a and <DCA=2a, DA=DC
    Because BD=EC, DC=DE+EC=DE+BD=BE, or DC=BE
    Note that <BAE=<BAD+<DAE=2a+a=3a, and <BEA=<EAC+<ECA=a+2a=3a, or <BAE=<BEA=3a, BE=BA
    From DA=DC, DC=BE, and BE=BA, we see that DA=BA.
    <BDA=<DAC+<DCA=2a+2a=4a
    In isosceles triangle BAD, <BDA=<DBA=4a and <BAD=2a.
    Therefore <BDA+<DBA+<BAD=4a+4a+2a=180, or a=18.

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