MathDB
Similar Triangles

Source: AHSME 1963 Problem 18

January 9, 2014
Asymptotegeometryperpendicular bisectorAMC

Problem Statement

Chord EFEF is the perpendicular bisector of chord BCBC, intersecting it in MM. Between BB and MM point UU is taken, and EUEU extended meets the circle in AA. Then, for any selection of UU, as described, triangle EUMEUM is similar to triangle:
[asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pair F = (0, 1); pair M = midpoint(B--C); pair A = (-0.99, -0.141); pair U = intersectionpoints(A--E, B--C)[0]; draw(B--C); draw(F--E--A); draw(unitcircle); label("BB", B, SW); label("CC", C, SE); label("AA", A, W); label("EE", E, S); label("UU", U, NE); label("MM", M, NE); label("FF", F, N); //Credit to MSTang for the asymptote [/asy]

<spanclass=latexbold>(A)</span> EFA<spanclass=latexbold>(B)</span> EFC<spanclass=latexbold>(C)</span> ABM<spanclass=latexbold>(D)</span> ABU<spanclass=latexbold>(E)</span> FMC<span class='latex-bold'>(A)</span>\ EFA \qquad <span class='latex-bold'>(B)</span>\ EFC \qquad <span class='latex-bold'>(C)</span>\ ABM \qquad <span class='latex-bold'>(D)</span>\ ABU \qquad <span class='latex-bold'>(E)</span>\ FMC