## Sunday, September 8, 2013

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

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

This entry contributed by Ajit Athle.
Click the figure below to see the complete problem 921. 1. Let ∠BAC = 2a, ∠BCA = 2c, ∴a+c=45
Note 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.

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

3. According to this result:
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.

4. Problem 921
Let 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

5. Good problem Ajith

Easily DIC bisects < C and EIA <A.

< AID = 45 = < AED (since < BDF = A/2 and < DBF = 45-A/2)

Hence ADFI is concyclic and so < AFI = < ADC = 90.
Similarly < IGC = 90 and the result follows.

Sumith Peiris
Moratuwa
Sri Lanka

6. Join D and E to O, from F and G draw perpendicular to AB, BC intersect at I
Right isoceles tr DOE similar to FIG