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1952 AMC 12 Question 19

AMC12 · 1952 AMC 12

Question 19

Angle $B$ of triangle $ABC$ is trisected by $BD$ and $BE,$ which meet $AC$ at $D$ and $E,$ respectively. Then:
  1. $\dfrac{AD}{EC}=\dfrac{AE}{DC}$
  2. $\dfrac{AD}{EC}=\dfrac{AB}{BC}$
  3. $\dfrac{AD}{EC}=\dfrac{BD}{BE}$
  4. $\dfrac{AD}{EC}=\dfrac{AB\cdot BD}{BE\cdot
    BC}$
  5. $\dfrac{AD}{EC}=\dfrac{AE\cdot BD}{DC\cdot BE}$