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

Labels:
120,
angle,
angle bisector,
degree,
quadrilateral,
triangle

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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