Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 698.
Monday, December 5, 2011
Problem 698: Quadrilateral, Diagonal, Angle Bisector, Triangle, 120 Degrees
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Let AB and DC intersect at P. Since m(APD)=60 and m(APD)+m(BEC)=180 PBEC is cyclic quadrilateral. PE is angle bisector of BAC. So, m(BEC)=m(EPB)=30
ReplyDeletehttp://img840.imageshack.us/img840/4838/problem698.png
ReplyDeleteExtend AB and CD to F ( See picture)
Note that angle(A)+ Angle (D)=120 and angle (AFD)=60
Since AE and EC are angle bisectors of angles A and C so FE is angle bisector of angle ( AFD)
Quadrilateral FBEC is cyclic ( F supplement to E) so x= angle (BFE)=30
Peter Tran
Problem 698 revisited:
ReplyDeleteWith F on AD let EF bisect angle AED.
Clearly
EA bisects angle BEF and
ED bisects angle CEF
So triangles AEB and AEF are congruent. So BE = EF.
Similarly CE = EF.
Hence BE = CE, triangle EBC is isosceles
with vertical angle at E = 120 deg
Hence each base angle = x = 30 deg
wonderfully innovative solution
DeleteThe solution is uploaded to the following link:
ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJbG9wSXBldmkweFE