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.

Let P be a point on AC such that AP=AB

ReplyDeleteLet 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