Geometry Problem. Post your solution in the comments box below.

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

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## Wednesday, December 24, 2014

### Geometry Problem 1069: Circle, Angles, Auxiliary Lines

Labels:
angle,
auxiliary line,
circle

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Join OB, then OB=OC and ∠OBC=x.

ReplyDeleteIn 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.png

ReplyDeleteDraw 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.

ReplyDeleteHence Right Tr.s OPB & OQC are congruent hence < OBP = < OCQ. So OABC is cyclic and x=70

Sumith Peiris

Moratuwa

Sri Lanka

https://photos.app.goo.gl/2Mn7PbRzd8BHYc8cA

ReplyDelete