Problem
1963 AHSME 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)[0]; 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]](/static/aops/a60944d306f088a476d6be2e167d8b32911f1fa1.png)
\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \dfrac{5}{2}
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