Geometry Problem

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

Click the figure below to see the complete problem 770.

## Thursday, June 21, 2012

### Problem 770: Parallelogram, Angle Bisector, Triangle, Circumcenter, Circle, Circumcircle, Angle Measurement

Labels:
angle bisector,
circumcenter,
circumcircle,
parallelogram,
triangle

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http://img542.imageshack.us/img542/8173/problem770.png

ReplyDeleteDraw lines per attached sketch

Note that ADE, CEF and ABF are isosceles triangles.

And BO⊥CE , ∠ECO=∠CEO= 90- α

AD=BC=DE

∆ OCB congruence to ∆ODE … ( Case SAS)

So OB=OD and triangle BOD is isosceles

X= (180-72)/2=54

To Peter Tran:

ReplyDeletePlease explain why

BO ⊥ CE

Where it is used?

Pravin

http://img62.imageshack.us/img62/9149/problem7701.png

DeletePravin

Thank you for your comment. The correct statement should be CO ⊥EF instead of BO⊥ CE . The solution and the result is valid with this correction. See attached sketch above for details

http://www.mathematica.gr/forum/viewtopic.php?f=20&t=29034&p=139447

ReplyDelete