Let's analyze the given problem involving triangle ABC and an extension of its side BC to a point D, creating similar triangles DAB and DCA.

Understanding the Problem with Similar Triangles

We are given a triangle ABC with the following side lengths:

• AB = 20 cm
• BC = 7 cm
• CA = 15 cm

Side BC is extended to a point D. This creates a new line segment BD. The problem states that triangle DAB is similar to triangle DCA (\(\Delta \text{DAB} \sim \Delta \text{DCA}\)). We need to find the length of the segment DC.

Properties of Similar Triangles

When two triangles are similar, their corresponding angles are equal, and the ratio of their corresponding sides is constant. The similarity statement \(\Delta \text{DAB} \sim \Delta \text{DCA}\) tells us which vertices correspond:

• D corresponds to D
• A corresponds to C
• B corresponds to A

This correspondence allows us to set up proportions between the lengths of the corresponding sides.

Setting Up Proportions

Based on the similarity \(\Delta \text{DAB} \sim \Delta \text{DCA}\), the ratios of corresponding sides are equal:

\(\frac{\text{DA}}{\text{DC}} = \frac{\text{AB}}{\text{CA}} = \frac{\text{DB}}{\text{DA}}\)

Let's denote the length of DC as \(x\) cm. Since BC is produced to D, the length of DB is the sum of BC and DC. Thus, DB = BC + DC = \(7 + x\) cm.

Now, substitute the known values into the proportions:

\(\frac{\text{DA}}{x} = \frac{20}{15} = \frac{7+x}{\text{DA}}\)

Solving for DC (x)

We have two useful equations from the proportions:

1. \(\frac{\text{DA}}{x} = \frac{20}{15}\)
2. \(\frac{20}{15} = \frac{7+x}{\text{DA}}\)

Let's simplify the ratio \(\frac{20}{15}\):

\(\frac{20}{15} = \frac{4 \times 5}{3 \times 5} = \frac{4}{3}\)

So the proportions become:

1. \(\frac{\text{DA}}{x} = \frac{4}{3}\)
2. \(\frac{4}{3} = \frac{7+x}{\text{DA}}\)

From equation (1), we can express DA in terms of \(x\):

\(\text{DA} = \frac{4}{3}x\)

Now, substitute this expression for DA into equation (2):

\(\frac{4}{3} = \frac{7+x}{\frac{4}{3}x}\)

To solve for \(x\), multiply both sides by \(\frac{4}{3}x\):

\(\frac{4}{3} \times \left(\frac{4}{3}x\right) = 7+x\)

\(\frac{16}{9}x = 7+x\)

Subtract \(x\) from both sides:

\(\frac{16}{9}x - x = 7\)

Combine the terms with \(x\):

\(\left(\frac{16}{9} - 1\right)x = 7\)

\(\left(\frac{16-9}{9}\right)x = 7\)

\(\frac{7}{9}x = 7\)

Multiply both sides by \(\frac{9}{7}\) to isolate \(x\):

\(x = 7 \times \frac{9}{7}\)

\(x = 9\)

The length of DC is 9 cm.

From \(\frac{\text{DA}}{x} = \frac{4}{3}\), we get \(\text{DA} = \frac{4}{3}x\).

Substituting DA in \(\frac{4}{3} = \frac{7+x}{\text{DA}}\):

\(\frac{4}{3} = \frac{7+x}{\frac{4}{3}x}\)

\(\frac{16}{9}x = 7+x\)

\(\left(\frac{16}{9} - 1\right)x = 7\)

\(\frac{7}{9}x = 7\)

\(x = 9\)

Conclusion

The length of DC is 9 cm.

Revision Table: Key Concepts

Additional Information: Applying Similar Triangle Properties

Similar triangles are a powerful tool in geometry for finding unknown lengths or angles. The key is correctly identifying the corresponding vertices and sides based on the similarity statement (e.g., \(\Delta \text{DAB} \sim \Delta \text{DCA}\)). The order of the vertices in the similarity statement is crucial.

In this problem, the vertex D is common to both triangles DAB and DCA. Angle DAB corresponds to angle DCA, and angle ABD corresponds to angle CAD. Angle ADB is the same as angle CDA (since it's the same angle).

The proportions can be derived directly from the vertex order:

First two vertices: DA / DC

Second two vertices: AB / CA

First and third vertices: DB / DA

Ensuring these ratios are set equal allows us to solve for any single unknown length, provided we have enough known lengths to form an equation.