MathDB
Angle bisectors

Source: 1959 AHSME Problem 28

August 14, 2013
AMCangle bisectorAMC 12geometry

Problem Statement

In triangle ABCABC, ALAL bisects angle AA and CMCM bisects angle CC. Points LL and MM are on BCBC and ABAB, respectively. The sides of triangle ABCABC are a,b,a,b, and cc. Then AMMB=kCLLB\frac{\overline{AM}}{\overline{MB}}=k\frac{\overline{CL}}{\overline{LB}} where kk is: <spanclass=latexbold>(A)</span> 1<spanclass=latexbold>(B)</span> bca2<spanclass=latexbold>(C)</span> a2bc<spanclass=latexbold>(D)</span> cb<spanclass=latexbold>(E)</span> ca <span class='latex-bold'>(A)</span>\ 1 \qquad<span class='latex-bold'>(B)</span>\ \frac{bc}{a^2}\qquad<span class='latex-bold'>(C)</span>\ \frac{a^2}{bc}\qquad<span class='latex-bold'>(D)</span>\ \frac{c}{b}\qquad<span class='latex-bold'>(E)</span>\ \frac{c}{a}