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.

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## Friday, November 6, 2015

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

Labels:
circle,
circumcenter,
congruence,
double angle,
parallel,
triangle

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Let a=18

ReplyDeleteBecause 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

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

ReplyDeleteSorry, my full proof must have been cut somehow

ReplyDeleteLet <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.