Problem: 半径为$$2$$的圆的直径$$AB$$被延伸到圆外的一点$$D$$，使$$BD=3$$.选择$$E$$点，使$$ED=5$$，直线$$ED$$与直线$$AD$$垂直．线段$$AE$$与圆相交于$$A$$和$$E$$之间的点$$C$$.请问三角形$$ABC$$的面积是多少？
Options:
aoVal: A
content: $$\frac{120}{37}$$ 
aoVal: B
content: $$\frac{140}{39}$$ 
aoVal: C
content: $$\frac{145}{39}$$ 
aoVal: D
content: $$\frac{140}{37}$$ 
aoVal: E
content: $$\frac{120}{31}$$ 
Answer: D
Analysis: Solution 1  Notice that $$ADE$$ and $$ABC$$ are right triangles.  Then $$AE=\sqrt{7^{2}+5^{2}}=\sqrt{74}$$,$$\sin DAE=\sin ABC=\dfrac{5}{\sqrt{74}}=\sin BAE=\dfrac{BC}{4}$$,so $$BC=\dfrac{20}{\sqrt{74}}$$. We also find that $$AC=\dfrac{28}{\sqrt{74}}$$, and thus the area of $$ABC$$ is $$\dfrac{\dfrac{20}{\sqrt{74}}\cdot\dfrac{28}{\sqrt{74}}}{2}=\dfrac{\dfrac{560}{74}}{2}=\dfrac{140}{37}$$.  Solution 2  We note that $$\triangle ACB\sim\triangle ADE$$ by $$AA$$ similarity. Also, since the area of $$\triangle ADE=\dfrac{7\cdot5}{2}=\dfrac{35}{2}$$ and $$AE=\sqrt {74}$$,$$\dfrac{\left[ ABC\right]}{\left[ ADE\right]}=\dfrac{\left[ ABC\right]}{\dfrac{35}{2}}=\left( \dfrac{4}{\sqrt{74}}\right)^{2}$$ , so the area of $$\triangle ABC=\frac {140}{37}$$.  Solution 3  As stated before, note that $$\triangle ACB$$,$$\triangle ADE$$. By similarity, we note that $$\dfrac{\overline{AC}}{\overline{BC}}$$ is equivalent to $$\frac 75$$. We set $$\overline{AC}$$ to $$7x$$~ and $$\overline{BC}$$ to $$5x$$. By the Pythagorean Theorem, $$\left( 7x\right)^{2}+\left( 5x\right)^{2}=4^{2}$$. Combining, $$49x^{2}+25x^{2}=16$$. We can add and divide to get $$x^{2}=\frac 8{37}$$. We square root and rearrange to get $$x=\dfrac{2\sqrt{74}}{37}$$. We know that the legs of the triangle are $$7x$$ and $$5x$$. Mulitplying $$x$$ by $$7$$ and $$5$$ eventually gives us $$\dfrac{14\sqrt{74}}{37}\dfrac{10\sqrt{74}}{37}$$. We divide this by $$2$$, since $$\frac 12bh$$ is the formula for a triangle. This gives us $$\frac {140}{37}$$.  Solution 4  Let\textquotesingle s call the center of the circle that segment $$AB$$ is the diameter of,$$O$$. Note that $$\triangle ODE$$ is an isosceles right triangle. Solving for side $$OE$$, using the Pythagorean theorem, we find it to be $$5\sqrt 2$$. Calling the point where segment $$OE$$ intersects circle $$O$$, the point $$I$$, segment $$IE$$ would be $$5\sqrt 2-2$$. Also, noting that $$\triangle ADE$$ is a right triangle, we solve for side $$AE$$, using the Pythagorean Theorem, and get $$\sqrt {74}$$. Using Power of Point on point $$E$$, we can solve for $$CE$$. We can subtract $$CE$$ from $$AE$$ to find $$AC$$ and then solve for $$CB$$ using Pythagorean theorem once more $$(AE)(CE)$$. $$=$$ (Diameter of circle $$O+IE$$ ),$$ \left( IE\right)\rightarrow\sqrt{74}\left( CE\right)=\left( 5\sqrt{2}+2\right)\left( 5\sqrt{2}-2\right)\Rightarrow CE=\dfrac{23\sqrt{74}}{37}$$,$$AC=AE-CE\rightarrow AC=\sqrt{74}-\dfrac{23\sqrt{74}}{37}\Rightarrow AC=\dfrac{14\sqrt{74}}{37}$$,Now to solve for $$CB$$:$$AB^{2}-AC^{2}=CB^{2}\rightarrow4^{2}+\dfrac{14\sqrt{74}^{2}}{37}=CB^{2}\Rightarrow CB=\dfrac{10\sqrt{74}}{37}$$ ,  Note that $$\triangle ABC$$ is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases $$AC$$ and $$BC$$, we get the area of triangle $$ABC$$ to be $$\frac {140}{37}$$. 
Knowledge points: 课内体系->知识点->直线和圆的方程
美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Circle
Source: 2017年AMC12竞赛B第18题