In triangle $ABC$, points $D$ and $E$ lie on the interior of segments $AB$ and $AC$, respectively,such that $AD = 1$, $DB = 2$, $BC = 4$, $CE = 2$ and $EA = 3$. Let $DE$ intersect $BC$ at $F$. Determine the length of $CF$.
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Tags: geometry
Imajinary
11.01.2021 02:09
2
through noticing the triangle is right and using LOC on CEF. Can someone check?
natmath
11.01.2021 02:41
$B$ is the origin, $A$ is $(3,0)$, and $B$ is $(0,4)$. By similar triangles, $E$ is $(\frac{6}{5},\frac{12}{5}$). $D$ is $(2,0)$. Again by similar triangles, $F$ is $(0,6)$. So $CF=2$
mathstudent5
21.12.2023 18:25
Use Menelaus theorem
resources
21.12.2023 19:37
let $CF=x$ by Menelaus's theorem, $\frac{1}{2} \cdot \frac{4+x}{x} \cdot \frac{2}{3} = 1$, so $CF=x=2$