Online Geometry theorems, problems, solutions, and related topics.
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.
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°.
http://s7.postimg.org/hjetqdsnv/Pro_1069.pngDraw 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.
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=70Sumith PeirisMoratuwaSri Lanka