It looks like the diagram represents a trapezoid, where $ED \parallel CB$, and we are tasked with finding the lengths of $AD$ and $DB$. To solve this, we can use the properties of similar triangles created by the parallel sides.

We can break the trapezoid into triangles, using the fact that the triangles $ADE$ and $CBE$ are similar due to the parallel lines. The side lengths along the parallel lines can help set up ratios for the lengths we need.

Step-by-Step Approach:

1. Label Known Values:

   • $AB = 75 \, m$
   • $EC = 25 \, m$
   • $AC = 30 \, m$
   • $ED \parallel CB$

2. Apply Properties of Similar Triangles:

   • Since $ED \parallel CB$, triangles $ADE$ and $CBE$ are similar, meaning their corresponding sides are proportional.

   Thus, we can write the proportion: $\frac{AD}{AC} = \frac{DB}{AB}$

3. Substitute the Known Values and Solve for $AD$ and $DB$: To solve for $AD$ and $DB$, we'll use the similar triangle properties and the known lengths. Let me do the math for you.

Let me calculate the values of $AD$ and $DB$. Known values AB = 75 # meters AC = 30 # meters EC = 25 # meters

Apply proportionality of similar triangles:

AD / AC = DB / AB -> AD = (AC * DB) / AB

First, we solve for the proportional parts.

AD = (EC * AB) / AC # Length of AD DB = AB - AD # Length of DB

AD, DBAfter calculating, the lengths are:

• $AD = 62.5 \, m$
• $DB = 12.5 \, m$

These are the lengths of segments $AD$ and $DB$, rounded to the nearest tenth.

Would you like more details or have any other questions about the solution? Here are some related questions for further understanding:

1. How do the properties of similar triangles help in solving this problem?
2. Can you apply the same method to other geometric shapes, such as parallelograms?
3. What happens if $ED$ is not parallel to $CB$?
4. How would you calculate the area of this trapezoid?
5. How do you use the Pythagorean theorem in problems involving right triangles?

Tip: When dealing with parallel lines and similar triangles, always remember that corresponding sides are proportional, which simplifies many geometry problems!