Problem:
In △ABC,D is on AC and F is on BC. Also, AB⊥AC,AF⊥BC, and BD=DC=FC=1. Find AC.
Answer Choices:
A. 2
B. 3
C. 32
D. 33
E. 43
Solution:
Let AC=x and ∠DCF=θ. Then ∠CBD=θ, and ∠ADB=2θ by the Exterior Angle Theorem. Thus cosθ=x1 and cos2θ=x−1. Therefore,
2(x1)2−12−x2x=x−1=x3−x2=32
OR
Drop DG⊥BC. Let AC=x,GC=y. Note that BC=2y, for △BDC is isosceles. Since △DCG∼△ACF∼△BCA, we obtain y1=1x=x2y. Thus y=x1 and y=2x2, implying x3=2, or x=32.
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