## Monday, December 5, 2011

### Problem 698: Quadrilateral, Diagonal, Angle Bisector, Triangle, 120 Degrees

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

Click the figure below to see the complete problem 698.

1. 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

2. http://img840.imageshack.us/img840/4838/problem698.png

Extend 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

3. Problem 698 revisited:
With 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

1. wonderfully innovative solution