Sunday, September 8, 2013

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

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.

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

4 comments:

  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.

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  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

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  3. According to this result:
    http://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.

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  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

    ReplyDelete