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# AE = ... .............?

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△ABC,

∠BAD = ∠DAE = ∠EAC,

BD = 2,

DE = EC = 1.

AE = ...

### 1 Answer

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- atsuoLv 64 years agoFavorite Answer
Let ∠BAD = ∠DAE = ∠EAC = θ .

Use the sine law ,

EC / sinθ = 1 / sinθ = AE / sin(∠ACE)

DE / sinθ = 1 / sinθ = AE / sin(∠ADE)

So

sin(∠ACE) = sin(∠ADE)

Both angles are smaller than 90° , so

∠ACE = ∠ADE

Therefore △ACD is an isosceles with AC = AD , so

∠AEC = ∠AED = 90° .

It means tan(2θ) = BE / AE = 3 / AE , so

2tanθ / (1 - tan^2(θ)) = 3 / AE

And tanθ = EC / AE = 1 / AE , so

2(1 / AE) / (1 - (1 / AE)^2) = 3 / AE

2AE / (AE^2 - 1) = 3 / AE

3(AE^2 - 1) = 2AE^2

AE^2 = 3

AE = √3 <--- The answer

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