Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.

Post your solution in the comments box below.

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

Click the figure below to see the dynamic geometry demostration of problem 887.

## Monday, June 10, 2013

### Problem 887: Triangle, Altitude, Angle Bisector, Perpendicular, Midpoint, Concyclic Points. GeoGebra

Labels:
altitude,
angle bisector,
animation,
concyclic,
GeoGebra,
HTML5,
midpoint,
perpendicular,
triangle

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http://img853.imageshack.us/img853/4498/problem887.png

ReplyDeleteLet CE intersect AB at F

Triangle BFC is isosceles => FE=EC

We have CE/EC= GC/GA => GE//AB

Since GE//AB => ∠(DEG)= ∠(ABE)

In cyclic quadrilateral ABDH we have ∠(DHG)= ∠(ABE)

So ∠(DEG)=∠(DHG) => H,D,G,E are concyclic

thanks peter tran

ReplyDeleteKey is to prove that DG//BC

ReplyDeleteThen < EDG = < EBC = < EHC = B/2 and the result follows