[Square](https://artofproblemsolving.com/wiki/index.php/Square) $ABCD$ has side length $2$. A [semicircle](https://artofproblemsolving.com/wiki/index.php/Semicircle) with [diameter](https://artofproblemsolving.com/wiki/index.php/Diameter) $\overline{AB}$ is constructed inside the square, and the [tangent](https://artofproblemsolving.com/wiki/index.php/Tangent_(geometry)) to the semicircle from $C$ intersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$?

$\mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad \mathrm{(B) \ } \sqrt{5} \qquad \mathrm{(C) \ } \sqrt{6} \qquad \mathrm{(D) \ } \frac{5}{2} \qquad \mathrm{(E) \ } 5-\sqrt{5}$
Solution 1

Let the point of tangency be $F$. By the [Two Tangent Theorem](https://artofproblemsolving.com/wiki/index.php/Two_Tangent_Theorem) $BC = FC = 2$ and $AE = EF = x$. Thus $DE = 2-x$. The [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) on $\triangle CDE$ yields
\begin{align*} DE^2 + CD^2 &= CE^2\\ (2-x)^2 + 2^2 &= (2+x)^2\\ x^2 - 4x + 8 &= x^2 + 4x + 4\\ x &= \frac{1}{2}\end{align*}
Hence $CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.

Solution 2
Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF$. Thus, $\angle EGF=\angle FCG$ and $\tan EGF=\tan FCG=\frac{1}{2}$. Solving $EF=\frac{1}{2}$. Adding, the answer is $\frac{5}{2}$.

Solution 3
[2004 AMC12A-18.png](https://artofproblemsolving.com/wiki/index.php/File:2004_AMC12A-18.png)
Clearly, $EA = EF = BG$. Thus, the sides of [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) $CDE$ are in arithmetic progression. Thus it is [similar](https://artofproblemsolving.com/wiki/index.php/Similar_triangles) to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.

Solution 4

Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\frac{1}{2}$, so the answer is $\frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.

Solution 5
Using the diagram as drawn in Solution 4, let the total area of square $ABCD$ be divided into the triangles $DCE$, $EAG$, $CGB$, and $EGC$. Let x be the length of AE. Thus, the area of each triangle can be determined as follows: 
\[DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = 2-x\]
\[EAG= \frac{AE\cdot{AG}}{2} = \frac{1\cdot{x}}{2} = \frac{x}{2}\]
\[CGB = \frac{GB\cdot{CB}}{2} = \frac{1\cdot(2)}{2} = 1\]
\[EGC= \frac{EG\cdot{GC}}{2} = \frac{\sqrt{5x^2 + 5}}{2}\] (the length of CE is calculated with the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), lines GE and 
CE are perpendicular by definition of tangent)
Adding up the areas and equating to the area of the total square \[(2 \cdot 2=4)\], we get
\[x = \frac{1}{2}\]
So, \[CE = 2 + \frac{1}{2} = 5/2\].
~Typo Fix by doulai1

Video Solution
[https://youtu.be/pM0zICtH6Lg](https://artofproblemsolving.comhttps://youtu.be/pM0zICtH6Lg)
Education, the Study of Everything