Geometry Problem. Post your solution in the comments box below.

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

This entry contributed by Ajit Athle.

Click the figure below to see the complete problem 921.

## Sunday, September 8, 2013

### Geometry Problem 921: Right Triangle, Circumcircle, Arc, Midpoint, Tangency Point, Incircle

Labels:
arc,
circumcircle,
incircle,
midpoint,
right triangle,
tangency point

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Let ∠BAC = 2a, ∠BCA = 2c, ∴a+c=45

ReplyDeleteNote that CD and AE are angle bisector of triangle ABC, so CD&AE = I

∠GEI=∠DEA=∠DCA=c=∠DCB=∠ICG, so G,I,E,C concylic,

∴∠GIE=∠GCE=∠BCE=∠BAE=a

∴∠FGI=∠GIE+∠GEI=a+c=45

∠BGF=∠EGC

=180-∠GEC-∠GCE

=180-(∠GEI+∠IEC)-a

=180-(c+90)-a

=45

Combining the above results,

∠BGI=∠BGF+∠FGI=45+45=90

∴G is the tangency point of incircle.

Symmetrical for F.

Q.E.D.

arch BE = arch EC => AE is the bisector of CD is the bisector of <BCA

ReplyDeleteAE ∩ CD = I, <AIC= 135 degrees, from this <CIE = 45 degrees.

<CGE = 1/2(arch AD + arch EC) = 45 degrees

G, E, C, I are concyclic.

<IGC = 90 degrees = <IEC

In an analogous manner <IFA = 90 degrees, this means F and G are the tangent points of the inscribed circle with center I.

Erina-NJ

According to this result:

ReplyDeletehttp://www.youtube.com/watch?v=CdpUpA5VxvA

Plus the well known fact that the angle bisector and the perpendicular bisector of the opposite side concur on the circumcircle and by taking into acount that the angle bisector is the middle paralel, we get trivially the dessired result.

Problem 921

ReplyDeleteLet DC intersect the AE at I(incenter).Then <IDF=<CDE=<CAE=<EAB=<FAI so

ADFI is cyclic ,then <AFI=<ADI=90 .Similar <IGC=<IEC=90. If IK is perpendicular at AC

Then IK=IF=IG therefore F and G are the points of tangency of AB and BC with the

Incircle I.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE