## Friday, November 6, 2015

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

Level: Mathematics Education, High School, Honors Geometry, College.

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

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

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.
<BDA=<DAC+<DCA=2a+2a=4a

4. Let P be a point on AC such that AP=AB
Since AF=FE => m(CFE)=2a=m(FCE) (since FE=EC)
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