Understanding the Problem: Triangle Area Division

The problem involves a triangle ABC with a line segment DE drawn parallel to the base BC. Point D is on side AB, and point E is on side AC. This creates a smaller triangle ADE within the larger triangle ABC. We are told that the line segment DE divides the area of the original triangle ABC into two equal parts. This means the area of triangle ADE is half the area of triangle ABC, and consequently, the area of the trapezoid DECB is also half the area of triangle ABC.

We need to find the ratio of the length of segment DB to the length of segment AB (DB : AB).

Applying Similar Triangle Properties

When a line segment is drawn parallel to one side of a triangle intersecting the other two sides, it creates a smaller triangle that is similar to the original triangle. In this case, since DE is parallel to BC, triangle ADE is similar to triangle ABC.

Similarity of triangles means that their corresponding angles are equal and their corresponding sides are proportional. A key property relating the areas of similar triangles is that the ratio of their areas is equal to the square of the ratio of their corresponding sides.

For similar triangles ADE and ABC, the ratio of their areas is:

\[ \frac{\text{Area}(\Delta \text{ADE})}{\text{Area}(\Delta \text{ABC})} = \left( \frac{\text{AD}}{\text{AB}} \right)^2 = \left( \frac{\text{AE}}{\text{AC}} \right)^2 = \left( \frac{\text{DE}}{\text{BC}} \right)^2 \]

Using the Area Information

We are given that the line segment DE divides the area of triangle ABC into two equal parts. This means:

\[ \text{Area}(\Delta \text{ADE}) = \frac{1}{2} \times \text{Area}(\Delta \text{ABC}) \]

Now, we can use the ratio of areas property:

\[ \frac{\text{Area}(\Delta \text{ADE})}{\text{Area}(\Delta \text{ABC})} = \frac{1}{2} \]

Equating this to the square of the ratio of corresponding sides (we'll use AD and AB):

\[ \left( \frac{\text{AD}}{\text{AB}} \right)^2 = \frac{1}{2} \]

Finding the Ratio AD : AB

To find the ratio AD : AB, we take the square root of both sides of the equation:

\[ \frac{\text{AD}}{\text{AB}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{1}}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

To rationalize the denominator, we multiply the numerator and denominator by \( \sqrt{2} \):

\[ \frac{\text{AD}}{\text{AB}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

So, the ratio AD : AB is \( \sqrt{2} : 2 \), or more commonly written as \( 1 : \sqrt{2} \).

Calculating the Ratio DB : AB

We are asked to find the ratio DB : AB. From the figure (or the problem description), point D lies on segment AB. This means that the length of AB is the sum of the lengths of AD and DB:

\[ \text{AB} = \text{AD} + \text{DB} \]

We can rearrange this equation to express DB in terms of AB and AD:

\[ \text{DB} = \text{AB} - \text{AD} \]

Now, let's find the ratio DB : AB by dividing both sides by AB:

\[ \frac{\text{DB}}{\text{AB}} = \frac{\text{AB} - \text{AD}}{\text{AB}} = \frac{\text{AB}}{\text{AB}} - \frac{\text{AD}}{\text{AB}} = 1 - \frac{\text{AD}}{\text{AB}} \]

We already found that \( \frac{\text{AD}}{\text{AB}} = \frac{1}{\sqrt{2}} \). Substituting this value:

\[ \frac{\text{DB}}{\text{AB}} = 1 - \frac{1}{\sqrt{2}} \]

To combine these terms, find a common denominator, which is \( \sqrt{2} \):

\[ \frac{\text{DB}}{\text{AB}} = \frac{\sqrt{2}}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \]

So, the ratio DB : AB is \( (\sqrt{2} - 1) : \sqrt{2} \).

Verification with Options

Let's compare our calculated ratio with the given options:

Our result, \( (\sqrt{2} - 1) : \sqrt{2} \), matches option 4.

Summary of Steps

1. Recognize that DE || BC implies \( \Delta \)ADE is similar to \( \Delta \)ABC.
2. Use the property that the ratio of areas of similar triangles is the square of the ratio of corresponding sides: \( \frac{\text{Area}(\Delta \text{ADE})}{\text{Area}(\Delta \text{ABC})} = \left( \frac{\text{AD}}{\text{AB}} \right)^2 \).
3. Use the given information that Area(\( \Delta \)ADE) = \(\frac{1}{2}\) Area(\( \Delta \)ABC), so \( \left( \frac{\text{AD}}{\text{AB}} \right)^2 = \frac{1}{2} \).
4. Solve for \( \frac{\text{AD}}{\text{AB}} \), getting \( \frac{\text{AD}}{\text{AB}} = \frac{1}{\sqrt{2}} \).
5. Express the desired ratio \( \frac{\text{DB}}{\text{AB}} \) in terms of \( \frac{\text{AD}}{\text{AB}} \): \( \frac{\text{DB}}{\text{AB}} = 1 - \frac{\text{AD}}{\text{AB}} \).
6. Substitute the value of \( \frac{\text{AD}}{\text{AB}} \) to find \( \frac{\text{DB}}{\text{AB}} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \).

Revision Table: Key Concepts

Additional Information: Proportionality Theorem

The fact that DE || BC implies \( \Delta \)ADE ~ \( \Delta \)ABC is a direct consequence of the Basic Proportionality Theorem (Thales's Theorem) and its converse. The theorem states that if a line parallel to one side of a triangle intersects the other two sides, it divides the two sides proportionally.

In our case, because DE || BC, we have:

\[ \frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}} \]

Also, it implies the proportionality of corresponding sides:

\[ \frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}} = \frac{\text{DE}}{\text{BC}} \]

This side proportionality is what allows us to use the area ratio property \( \left( \frac{\text{AD}}{\text{AB}} \right)^2 \).

Understanding the connection between parallel lines, similar triangles, side ratios, and area ratios is crucial for solving problems like this involving a triangle cut by a line parallel to its base.