Question 5, RMO 2003, issue with ratios

Question

In problem 5, RMO 2003 a specific part of the solution depends on the following $$\dfrac{BD}{DC} = \dfrac{AE}{EC} = \dfrac{AF}{FB} = \dfrac{DC}{BD}$$

It is proven that $AB \parallel DE \: , BC \parallel FE \:, AC \parallel FD \:\:\:$ So I can see that $$\Delta EDC \sim \Delta ABC \implies \dfrac{BC}{DC} - 1 = \dfrac{AC}{EC} - 1$$

But how come this ratio continues to the next line $FD$?

Answer 1

All equalities directly follow from the intercept theorem (take a look at https://en.wikipedia.org/wiki/Intercept_theorem) . You don't need to search for similar triangles.

• Equation $\dfrac{BD}{DC} = \dfrac{AE}{EC}$: $C$ is the intersection of $BC$ and $AC$. $AB$ is parallel to $ED$. So you can apply the intercept theorem.
• Equation $\dfrac{AE}{EC} = \dfrac{AF}{FB}$: $A$ is the intersection of $AC$ and $AB$. $FE$ is parallel to $BC$. So you can apply the intercept theorem.
• Equation $\dfrac{AF}{FB} = \dfrac{DC}{BD}$: $B$ is the intersection of $AB$ and $BC$. $FD$ is parallel to $AC$. So you can apply the intercept theorem.