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Source: 1971 AHSME Problem 26

April 23, 2014
ratioAMC

Problem Statement

[asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (3,0); E = (0,0); F = (1,2); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("AA",A,N); label("BB",B,W); label("CC",C,dir(0)); label("EE",E,S); label("FF",F,NE); label("GG",G,SE); //Credit to chezbgone2 for the diagram[/asy]
In triangle ABCABC, point FF divides side ACAC in the ratio 1:21:2. Let EE be the point of intersection of side BCBC and AGAG where GG is the midpoints of BFBF. The point EE divides side BCBC in the ratio
<spanclass=latexbold>(A)</span>1:4<spanclass=latexbold>(B)</span>1:3<spanclass=latexbold>(C)</span>2:5<spanclass=latexbold>(D)</span>4:11<spanclass=latexbold>(E)</span>3:8<span class='latex-bold'>(A) </span>1:4\qquad<span class='latex-bold'>(B) </span>1:3\qquad<span class='latex-bold'>(C) </span>2:5\qquad<span class='latex-bold'>(D) </span>4:11\qquad <span class='latex-bold'>(E) </span>3:8