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2 to 1 Bisector (Posted on 2014-03-30) Difficulty: 3 of 5

  
Let ∠ABC of ΔABC be a right angle and the bisector of angle ∠BAC
intersect the altitude BD at point E. Prove the following:

If m(∠BCE) = 2*m(∠ACE), then |CE| = 2*|BD|.
  

See The Solution Submitted by Bractals    
Rating: 4.0000 (1 votes)

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Some Thoughts A track. (right track?) | Comment 1 of 2
Let BD=1
Call ∠BAD=θ
So ∠EAD=θ/2 and to make the 'if' part of the statement true ∠ECD=30-θ/3 degrees

Need to show CE/BD= CE/1 = CE = 2
CE = ED/sin(30-θ/3)
CE = tan(θ/2)/[tan(θ)sin(30-θ/3)]

So really just need θ or tan(θ) or some trig function of θ to evaluate this.

Unfortunately the equations that come up don't solve easily.

AD = 1/tan(θ)
ED = ADtan(θ/2) = tan(θ/2)/tan(θ)
AD*DC=BDČ=1
DC=1/AD = tan(θ)

tan(∠ECD)=tan(30-θ/3) = ED/DC
tan(30-θ/3)=[tan(θ/2)/tan(θ)]/tan(θ)
and this is where I'm stuck.
θ≈63.15323 degrees but I can't find a closed form for it or any trig function of it.

  Posted by Jer on 2014-03-31 12:56:05
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