Tuesday, January 3, 2012

Problem 712: Semicircle, Diameter, Tangent, Perpendicular, Angle Measure

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

Click the figure below to see the problem 712 details.

Online Geometry Problem 712: Semicircle, Diameter, Tangent, Perpendicular, Angle Measure.

6 comments:

  1. http://img38.imageshack.us/img38/9066/problem712.png
    Connect O’D, BC . Draw O’E perpendicular to BC ( see picture)
    Note that O’D and BC perpen. To AC
    m(FDO’)=m(EO’B)=25
    Triangle FDO’ congruence to tri. EO’B so DF=O’E=DC
    Tri. FDC is isosceles and m(CFB)=1/2*m(FDC)= 57.5

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  2. Join BD, BC, O'D.
    W.r.t. circle (O'):
    Angle subtended by arc BD at centre O'
    = ∠ BO'D = 90° + 25° = 115°.
    So ∠ CDB = angle between tangent CB and chord DB =Half the angle subtended by arc DB at centre O' = (∠DO’B)/2 = 115°/2 = 57.5°
    But D,F,B,C are con-cyclic (Since ∠DFB + ∠DCB
    = 90° + 90° = 180°)
    Hence x = ∠ BFC = ∠ BDC = 57.5°

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  3. To Pravin: I think there's a little flaw in your proof. You say 'tangent CB', but I don't think CB is a tangent!
    You can repair this line of your proof with the following lines:
    O'D = O'B = radius, so triangle O'BD is isosceles and ∠O'DB = ∠O'BD = (180° - ∠BO'D)/2 = 32,5°
    ∠O'DC = 90° (tangent), so ∠CDB = 90°-∠O'DB = 57,5°

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  4. To Henkie:
    It is only a typo - I meant tangent CD.
    As such there is no flaw in the proof.
    You could as well take the angle between the (common)tangent BT at B, and BD.
    Thank you , anyway.
    Pravin.

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  5. Draw common tangent BG, G on AC.
    Tangents GD=GB hence < GDB = 57 1/2 which is = x since DFCB is cyclic.
    Sumith Peiris
    Moratuwa
    Sri Lanka

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  6. Problem 712
    Is <BFD=90=<DCB ,then D,C,B,F are concyclic. So <CFB=<FAC+<FCA=25+<FCD=25+<FBD=25+(<DO’A)/2=25+65/2=57.5.
    MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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