In the given figure, ∆ABC has points D and F in AC andpoint E in BC such that DE ∥ AB and EF ∥ BD. If CF = 4 cmand AC = 9 cm, what is the length of DC?

The problem involves similar triangles and the concept of proportionality.

Since DE is parallel to AB, triangle ADE is similar to triangle ABC (by the AA criterion).

Similarly, since EF is parallel to BD, triangle BCF is similar to triangle BDA.

From the similarity of triangles ADE and ABC, we have:

AD/AB = DE/BC = AE/AC

We don't have values for these lengths, so let's move on to the other set of similar triangles.

From the similarity of triangles BCF and BDA, we have:

BC/BD = CF/DA = BF/BA

We know that CF = 4 cm and AC = 9 cm.

Since CF/DA = AC/AD, we can set up the proportion 4/DA = 9/AD.

Cross-multiplying gives us 4AD = 9DA.

This equation tells us that AD = 9/4 DA.

But we also know that AD + DA = AC, or AD + AD*(9/4) = 9.

Solving this equation for AD, we get AD = 36/13 cm.

Therefore, DC = AC - AD = 9 - 36/13 = 117/13 - 36/13 = 81/13 cm, or approximately 6.23 cm.