AMC 12 Mock Exam 1 Problem 12

Problem 12 of 25

In triangle $ABC$ , $AB=10$ , $BC=12$ , and $CA=16$ . A circle passes through $A$ and $B$ and is tangent to line $BC$ at $B$ . The circle meets side $\overline{AC}$ again at a point $D$ other than $A$ . What is $BD$ ?

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Since line $BC$ is tangent to the circle at $B$ , the power of the point $C$ gives $CB^2=CD\cdot CA$ , so $CD=\frac{12^2}{16}=9$ . (As $CD<CA$ , the point $D$ lies on side $\overline{AC}$ .)

Now $\frac{CD}{CB}=\frac{9}{12}=\frac34=\frac{CB}{CA}$ and triangles $CDB$ and $CBA$ share the angle at $C$ , so $\triangle CDB\sim\triangle CBA$ with ratio $\frac34$ . Therefore $BD=\frac34\cdot AB=\frac34\cdot10=\boxed{\tfrac{15}{2}}$ .

Tagged: tangent circle, power of a point, similar triangles

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