Wednesday, December 24, 2014

Geometry Problem 1069: Circle, Angles, Auxiliary Lines

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1069: Circle, Angles, Auxiliary Lines.


  1. Join OB, then OB=OC and ∠OBC=x.

    In ΔOAB, OA/sin(∠OBA)=OB/sin110°
    In ΔOAC, OA/sin(∠OCA)=OC/sin70°

    Thus, ∠OBA = ∠OCA.
    Therefore, OABC is cyclic quadrilateral.

    Hence, x=70°.


    Draw circumcircle of triangle OAC and let AB cut circumcircle of OAC at B’
    In circle OAC we have Arc(OC)= 140 degrees, Arc(B’C)= 80 degrees so Arc(OAB’)=140
    So OC=OB’=OB => B coincide to B’
    Angle x= ½ Arc(OAB’)= 70 degrees.

  3. Draw perpendiculars OP & OQ to AB & AC respectively.Then Tr.s OAP & OAQ are congruent SAA. So OP=OQ.
    Hence Right Tr.s OPB & OQC are congruent hence < OBP = < OCQ. So OABC is cyclic and x=70

    Sumith Peiris
    Sri Lanka