The points D and E are taken on the sides `AB and AC `of `DeltaABC`such that `AD = 1/3 AB, AE = 1/3 AC`. If the length of BC is 15 cm, then the length of DE is :

Options

A. 10 cm
B. 8 cm
C. 6 cm
D. 5 cm

Answer

To find the length of segment DE in triangle ABC, where points D and E are located on sides AB and AC respectively, we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- \( AD = \frac{1}{3} AB \)
- \( AE = \frac{1}{3} AC \)
- \( BC = 15 \, \text{cm} \)

### Step 2: Set Up the Ratios
From the information provided, we can establish the following ratios:
- The ratio of \( AD \) to \( AB \) is \( \frac{AD}{AB} = \frac{1}{3} \)
- The ratio of \( AE \) to \( AC \) is \( \frac{AE}{AC} = \frac{1}{3} \)

### Step 3: Apply the Basic Proportionality Theorem (Thales' Theorem)
According to Thales' theorem, if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. Therefore, we can say:
\[ \frac{DE}{BC} = \frac{AD}{AB} = \frac{AE}{AC} \]
Since both ratios are equal to \( \frac{1}{3} \), we can express this as:
\[ \frac{DE}{15} = \frac{1}{3} \]

### Step 4: Solve for DE
To find DE, we can rearrange the equation:
\[ DE = \frac{1}{3} \times 15 \]
Calculating this gives:
\[ DE = 5 \, \text{cm} \]

### Conclusion
The length of segment DE is \( 5 \, \text{cm} \).

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