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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 712 details.
http://img38.imageshack.us/img38/9066/problem712.pngConnect O’D, BC . Draw O’E perpendicular to BC ( see picture)Note that O’D and BC perpen. To ACm(FDO’)=m(EO’B)=25Triangle FDO’ congruence to tri. EO’B so DF=O’E=DCTri. FDC is isosceles and m(CFB)=1/2*m(FDC)= 57.5
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°
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°
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.
Draw common tangent BG, G on AC.Tangents GD=GB hence < GDB = 57 1/2 which is = x since DFCB is cyclic.Sumith PeirisMoratuwaSri Lanka
Problem 712Is <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