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# Understand the problem

A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle? $[asy]draw((0,0)--(8,15)--(16,0)--(0,0)); draw(arc((8,0),7.0588,0,180));[/asy]$ $\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}$
##### Source of the problem
2016 AMC (American Mathematical Contests) 8  problem 25
Number Theory
##### Difficulty Level
Easy

Do you really need a hint? Try it first!

Hint figure :   $[asy] pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,S); label("E",E,NW);[/asy]$
$[asy] pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,S); label("E",E,NW);[/asy]$  Verify $$AD = 8$$ and recall the concepts of Similar Triangles. Then proceed .
$[asy] pair A, B, C, D, E; A=(0,0); B=(16,0); C=(8,15); D=B/2; E=(64/17*8/17, 64/17*15/17); draw(A--B--C--cycle); draw(C--D); draw(D--E); draw(arc(D,120/17,0,180)); draw(rightanglemark(B,D,C,25)); draw(rightanglemark(A,E,D,25)); label("A",A,SW); label("B",B,SE); label("C",C,N); label("D",D,S); label("E",E,NW);[/asy]$   Verify $$\triangle AED \sim \triangle ACD$$ . And proceed to find the radius of the semicircle i.e. DE . [NOTE : As DE is the radius and AC be the tangent so $$DE \perp AC$$  .]
We have $$\triangle AED \sim \triangle ACD$$ . This similarity means that we can create a proportion: $$\frac {AD}{AC} = \frac {DE}{CD}$$ . We plug in $$AD = 8 , AC = 17 \ and \ CD = 15$$ . So $$DE = \frac {8}{17} \times 15 = \frac {120}{17}$$  .

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