# Difference between revisions of "1989 AHSME Problems/Problem 15"

## Problem

In $\triangle ABC$, $AB=5$, $BC=7$, $AC=9$, and $D$ is on $\overline{AC}$ with $BD=5$. Find the ratio of $AD:DC$.

$[asy] draw((3,4)--(0,0)--(9,0)--(3,4)--(6,0)); dot((0,0));dot((9,0));dot((3,4));dot((6,0)); label("A", (0,0), W); label("B", (3,4), N); label("C", (9,0), E); label("D", (6,0), S);[/asy]$

$\textrm{(A)}\ 4:3\qquad\textrm{(B)}\ 7:5\qquad\textrm{(C)}\ 11:6\qquad\textrm{(D)}\ 13:5\qquad\textrm{(E)}\ 19:8$

## Solution

$[asy] draw((3,4)--(0,0)--(9,0)--(3,4)--(6,0)); draw((3,4)--(3,0),dotted); dot((0,0));dot((9,0));dot((3,4));dot((6,0));//dot((3,0)); label("A", (0,0), W); label("B", (3,4), N); label("C", (9,0), E); label("D", (6,0), S); label("h", (3,1.5),W); draw(rightanglemark((3,1),(3,0),(4,0),10)); label("x",(1.5,0),S);label("x",(4.5,0),S); [/asy]$ Drop the altitude $h$ from $B$ through $AD$, and let $AD$ be $2x$. Then by Pythagoras $$\begin{cases}h^2+x^2=25\\h^2+(9-x)^2=49\end{cases}$$ and after subtracting the first equation from the second, $x=3\tfrac16$. Therefore the desired ratio is $$\frac{6\tfrac13}{2\tfrac23}=\boxed{\frac{19}8}$$

## Solution 2 (Trig)

Using laws of cosines on $\bigtriangleup ABC$ yields $49=25+81-2 \cdot 5 \cdot 9 \cdot \cos A \implies \cos A =\frac{19}{3}.$ Let $AD=x.$ Using laws of cosines on $\bigtriangleup ABD$ yields $25+x^2-2 \cdot 5 \cdot x \cdot \cos A = 25.$ Fortunately, we know that $\cos A =\frac{19}{3}.$ Plugging this information back into our equation yields $x=\frac{19}{3}.$ Then, we know that $DC=9-\frac{19}{3}=\frac{8}{3} \implies\frac{AD}{DC}=\boxed{\frac{19}{8}}.$