I assume that after bisecting angle C, resulting in line segment CG (where G is the point where the bisector intercepts DB)[I used G because we already have used F in Nick's solution], the next step is to prove that either Angle GCB or Angle ECG is equal to angle BDE, which is known to be twice Angle A.
Either that, or show that Angle BGC is 3 times Angle A [and since it must also be A + 1/2 (ECB), then 1/2 (ECB) = 2(A), so ECB = 4(A)]
Whichever is the way you did it, I can't see how you established the next step.
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Posted by TomM
on 2002-06-20 17:07:41 |