Let's break down this geometry problem involving triangles, parallel lines, and ratios of areas. We are given a triangle ΔABC, with points D and E on sides AB and AC respectively. We know that the line segment DE is parallel to the side BC, and the ratio of the lengths AD to DB is 7:9. The line segments CD and BE intersect at point F. Our goal is to find the ratio of the area of ΔDEF to the area of ΔCBF.

Understanding Similar Triangles in the Figure

The fact that DE is parallel to BC is crucial. This parallelism creates similar triangles. Consider the triangle ΔABC and the line segment DE parallel to BC cutting sides AB and AC.

Since DE ∥ BC, by the Basic Proportionality Theorem (or Thales's Theorem) and its converse, we know that ΔADE is similar to ΔABC. The corresponding angles are equal (∠ADE = ∠ABC, ∠AED = ∠ACB, and ∠DAE = ∠BAC is common).

Also, the ratio of corresponding sides of similar triangles is equal. So, we have:

\begin{equation*} \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} \end{equation*}

Using the Given Ratio AD:DB

We are given the ratio AD ∶ DB = 7 ∶ 9. Let AD = 7x and DB = 9x for some positive value x. Then the total length of side AB is AD + DB = 7x + 9x = 16x.

Now we can find the ratio AD/AB:

\begin{equation*} \frac{AD}{AB} = \frac{7x}{16x} = \frac{7}{16} \end{equation*}

Since $\triangle ADE \sim \triangle ABC$, the ratio of corresponding sides is equal to AD/AB.

\begin{equation*} \frac{DE}{BC} = \frac{AD}{AB} = \frac{7}{16} \end{equation*}

This gives us the ratio of the lengths of the parallel segments DE and BC.

Finding Another Pair of Similar Triangles for Area Ratio

Now, consider the triangles ΔDEF and ΔCBF, which are formed by the intersection of CD and BE at F. Since DE ∥ BC, we can identify pairs of alternate interior angles:

• ∠EDF and ∠BCF (formed by transversal CD)
• ∠DEF and ∠CBF (formed by transversal BE)

Also, the angles ∠DFE and ∠BFC are vertically opposite angles, and thus they are equal.

Since all three pairs of corresponding angles are equal, ΔDEF is similar to ΔCBF by AAA similarity.

Calculating the Ratio of Areas of Similar Triangles

For any two similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides.

We have established that ΔDEF ∼ ΔCBF. The corresponding sides are DE and CB, DF and CF, EF and BF. We already found the ratio of the corresponding sides DE and BC (which is the same as CB):

\begin{equation*} \frac{DE}{CB} = \frac{7}{16} \end{equation*}

Therefore, the ratio of the area of ΔDEF to the area of ΔCBF is:

\begin{equation*} \frac{\text{Area}(\Delta DEF)}{\text{Area}(\Delta CBF)} = \left(\frac{DE}{CB}\right)^2 = \left(\frac{7}{16}\right)^2 \end{equation*}

Calculating the square:

\begin{equation*} \left(\frac{7}{16}\right)^2 = \frac{7^2}{16^2} = \frac{49}{256} \end{equation*}

So, the ratio of the areas of ΔDEF and ΔCBF is 49 ∶ 256.

Ratio of Areas Conclusion

The ratio of the areas of ΔDEF and ΔCBF is 49 ∶ 256.

Revision Table: Key Concepts for Triangle Area Ratios

Additional Information: Extending Triangle Properties

Understanding similar triangles and their area relationships is fundamental in geometry. The ratio of areas being the square of the ratio of corresponding sides is a powerful tool. Let's explore some related points:

• Ratio of Perimeters: For similar triangles, the ratio of their perimeters is simply equal to the ratio of their corresponding sides.
• Ratio of Altitudes/Medians/Angle Bisectors: For similar triangles, the ratio of corresponding altitudes, medians, or angle bisectors is also equal to the ratio of their corresponding sides.
• Areas of Triangles with the Same Height or Base:
  • If two triangles share the same height, the ratio of their areas is equal to the ratio of their bases. Area(Δ1)/Area(Δ2) = Base1/Base2.
  • If two triangles share the same base, the ratio of their areas is equal to the ratio of their corresponding heights. Area(Δ1)/Area(Δ2) = Height1/Height2.
• In problems like this, breaking down the complex figure into simpler triangles and identifying relationships (like similarity or shared height/base) is key to finding area ratios.

This problem specifically used the similarity property to relate the small triangle DEF and the larger triangle CBF based on the parallel lines DE and BC.