Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Find the answer

Example 1. As shown in the figure, $B D$ : $D C$ : $C E$ : $E A=2$ : $1, A D, B E$ intersect at $F$ , then $A F: F D=$ $\qquad$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

$\begin{array}{l} \because \lambda_{1}=\lambda_{2}=\frac{1}{2}, \\ \therefore \lambda_{3}=\left(1+\frac{1}{2}\right) \cdot \frac{1}{2}=\frac{3}{4} . \end{array}$

In this problem, $\triangle ABC$ only has the characteristics of a basic figure, and $D, F$ are known fixed points dividing $BC, AC$ respectively. Therefore, we can directly use (*) to solve.
$\begin{array}{l} \because \lambda_{1}=\lambda_{2}=\frac{1}{2}, \\ \therefore \lambda_{3}=\left(1+\frac{1}{2}\right) \cdot \frac{1}{2}=\frac{3}{4} . \end{array}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.