# 1963 AHSME Problems/Problem 39

## Problem 39

In $\triangle ABC$ lines $CE$ and $AD$ are drawn so that $\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$ where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals: $[asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P = intersectionpoints(A--D, C--E); draw(A--B--C--cycle); draw(A--D); draw(C--E); label("A", A, SW); label("B", B, SE); label("C", C, N); label("D", D, NE); label("E", E, S); label("P", P, S); //Credit to MSTang for the asymptote[/asy]$ $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \dfrac{5}{2}$

## Solution $[asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P = intersectionpoints(A--D, C--E); draw(A--B--C--cycle); draw(A--D); draw(C--E); label("A", A, SW); label("B", B, SE); label("C", C, N); label("D", D, NE); label("E", E, S); label("P", P, S); draw(P--B,dotted); //Credit to MSTang for the asymptote[/asy]$

Draw line $PB$, and let $[PEB] = 2b$, $[PDB] = a$, and $[CAP] = c$, so $[CPD] = 3a$ and $[APE] = 3b$. Because $\triangle CAE$ and $\triangle CEB$ share an altitude, $$c + 3b = \tfrac{3}{2} (3a+a+2b)$$ $$c + 3b = 6a + 3b$$ $$c = 6a$$ Because $\triangle ACD$ and $\triangle ABD$ share an altitude, $$6a+3a = 3(a+2b+3b)$$ $$9a = 3a+15b$$ $$6a = 15b$$ $$a = \tfrac{5}{2}b$$ Thus, $[CAP] = 15b$, and since $[APE] = 3b$, $r = \tfrac{CP}{PE} = 5$, which is answer choice $\boxed{\textbf{(D)}}$.

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