MathDB
Similar triangles

Source: AHSME 1986 problem 16

October 1, 2011
AMC

Problem Statement

In ABC\triangle ABC, AB=8AB = 8, BC=7BC = 7, CA=6CA = 6 and side BCBC is extended, as shown in the figure, to a point PP so that PAB\triangle PAB is similar to PCA\triangle PCA. The length of PCPC is
[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("AA", A, W); label("BB", B, E); label("CC", C, NE); label("PP", P, NW); label("66", 3*dir(A--C), SE); label("77", B+3*dir(B--C), NE); label("88", (4,0), S);[/asy] <spanclass=latexbold>(A)</span> 7<spanclass=latexbold>(B)</span> 8<spanclass=latexbold>(C)</span> 9<spanclass=latexbold>(D)</span> 10<spanclass=latexbold>(E)</span> 11 <span class='latex-bold'>(A)</span>\ 7\qquad<span class='latex-bold'>(B)</span>\ 8\qquad<span class='latex-bold'>(C)</span>\ 9\qquad<span class='latex-bold'>(D)</span>\ 10\qquad<span class='latex-bold'>(E)</span>\ 11