Try this beautiful problem from American Mathematics Competitions, AMC 10A, 2004 based on Geometry: Length of a Tangent.
Square \(ABCD\) has a side length \(2\). A Semicircle with diameter \(AB\) is constructed inside the square, and the tangent to the semicircle from \(C\) intersects side \(AD\)at \(E\). What is the length of \(CE\)?
Square
Semi-circle
Geometry
But try the problem first...
Answer: \(\frac{5}{2}\)
AMC-10A (2004) Problem 22
Pre College Mathematics
First hint
We have to find out length of \(CE\).Now \(CE\) is a tangent of inscribed the semi circle .Given that length of the side is \(2\).Let \(AE=x\).Therefore \(DE=2-x\). Now \(CE\) is the tangent of the semi-circle.Can you find out the length of \(CE\)?
Can you now finish the problem ..........
Second Hint
Since \(EC\) is tangent,\(\triangle COF\) \(\cong\) \(\triangle BOC\) and \(\triangle EOF\) \(\cong\) \(\triangle AOE\) (By R-H-S law).Therefore \(FC=2\) & \( EC=x\).Can you find out the length of \(EC\)?
can you finish the problem........
Final Step
Now the \(\triangle EDC\) is a Right-angle triangle........
Therefore \(ED^2+ DC^2=EC^2\) \(\Rightarrow (2-x)^2 + 2^2=(2+x)^2\) \(\Rightarrow x=\frac{1}{2}\)
Hence \(EC=EF+FC=2+\frac{1}{2}=\frac{5}{2}\)
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