1983 AHSME Problems/Problem 19

Problem

Point $D$ is on side $CB$ of triangle $ABC$ . If $\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6$ , then the length of $AD$ is

$\textbf{(A)} \ 2 \qquad \textbf{(B)} \ 2.5 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 3.5 \qquad \textbf{(E)} \ 4$

Solution 1 (Law of Cosines)

Let $AD = y$ . Since $AD$ bisects $\angle{BAC}$ , the Angle Bisector Theorem gives $\frac{DB}{CD} = \frac{AB}{AC} = 2$ , so let $CD = x$ and $DB = 2x$ . Applying the Law of Cosines to $\triangle CAD$ gives $x^2 = 3^2 + y^2 - 3y$ , and to $\triangle DAB$ gives $(2x)^2 = 6^2 + y^2 - 6y$ . Subtracting $4$ times the first equation from the second equation therefore yields $0 = 6y - 3y^2 \Rightarrow y(y-2) = 0$ , so $y$ is $0$ or $2$ . But since $y \neq 0$ ( $y$ is the length of a side of a triangle), $y$ must be $2$ , so the answer is $\boxed{\textbf{(A)} \ 2}$ .

Solution 2 (Similar Triangles)

Let $E$ be a point on line $AC$ such that $AE = 6$ and $A$ is between $E$ and $C$ . Then $\triangle ABE$ must be equilateral, $BE = 6$ , and $\angle BEC = 60^\circ$ . So $BE \parallel DA$ and $\triangle CDA \sim \triangle CBE$ . So $\frac{AD}{6} = \frac{3}{9}$ , so $AD = \boxed{(\mathbf{A})\ 2}$ .

-j314andrews

1983 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1983 AHSME Problems/Problem 19

Contents

Problem

Solution 1 (Law of Cosines)

Solution 2 (Similar Triangles)

See Also