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.


  1. Let a=18
    Because triangles ABE and AFC are both isosceles, BF perpendicularly bisects AE, meaning that
    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

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

  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.
    In isosceles triangle BAD, <BDA=<DBA=4a and <BAD=2a.
    Therefore <BDA+<DBA+<BAD=4a+4a+2a=180, or a=18.

  4. Let P be a point on AC such that AP=AB
    Let BD=AF=FE=EC=x and DE=y. Denote m(CAE)=m(EAD)=a => m(BAD)=2a (given)
    Triangles BAD and PAD are congruent (SAS), hence PD=BD=x -------------(1)
    Since AF=FE => m(CFE)=2a=m(FCE) (since FE=EC)
    Observe that m(CAD)=m(DCA)=2a
    => ADC is isosceles and AD=DC=x+y and hence m(ADB)=4a and m(ABC)=180-6a -------------(2)
    Now consider the triangle ABE, from (2) and given m(EAB)=3a, we have m(BEA)=3a => AB=BE=x+y -----(3)
    From (2)&(3), AD=AB=x+y => the triangle ABD is isosceles
    => m(ABD)=m(ADB)
    => 180-6a=4a
    => a=18 degrees ---------(4)

    Plugging the value of 'a' in the triangle ABC, we can see that it is a 72-72-36 isosceles triangle ----(5)
    From (3), AB=AP=x+y=>FP=y and from (5), AC=BC=2x+y=>PC=x and from the triangle CDF, CD=CF=x+y ----(6)
    From (6), we can say that FD//AB
    Join BF and consider the triangle BEF
    Observe that BE=AD=x+y, m(BEF)=m(ADP)=4a, EF=DP=x
    Hence the triangles BEF and ADP are congruent (SAS) and BEF is isosceles => BF=BE=BA => B is the circumcenter of triangle AEF