The problem asks for the length of the line segment $BD$ in the given quadrilateral $ABCD$, where:

• $AD = AB = 30$
• $CD = CB = 40$
• $\angle ADC = \angle ABC = 90^\circ$

This indicates that quadrilateral $ABCD$ consists of two right-angled triangles: $\triangle ADC$ and $\triangle ABC$. We will apply the Pythagorean theorem to each triangle.

Step 1: Applying the Pythagorean Theorem to $\triangle ADC$

In $\triangle ADC$, we have:

• $AD = 30$
• $CD = 40$
• $\angle ADC = 90^\circ$

Using the Pythagorean theorem: $AC^2 = AD^2 + CD^2$ $AC^2 = 30^2 + 40^2 = 900 + 1600 = 2500$ $AC = \sqrt{2500} = 50$

Step 2: Applying the Pythagorean Theorem to $\triangle ABC$

In $\triangle ABC$, we know:

• $AB = 30$
• $BC = 40$
• $AC = 50$

Now, we apply the Pythagorean theorem again to find the length of $BD$ (denoted as $z$). Since $BD$ is the diagonal across two right-angled triangles, we use the property of the diagonal in such cases, leading us to:

Since both triangles are identical, the length of $BD$ is twice the base diagonal, which confirms the choice closest to $50$.

Thus, the length of the line segment $BD$ is: $\boxed{50}$

Would you like further clarifications or have any additional questions?

Here are some related questions for further exploration:

1. What is the perimeter of quadrilateral $ABCD$?
2. How would you calculate the area of the entire quadrilateral $ABCD$?
3. Can you derive the diagonal $BD$ using vector algebra?
4. How would the problem change if one of the angles was not 90 degrees?
5. How can you use trigonometry to solve for $BD$?

Tip: Always check if right-angle triangles are present in geometry problems, as they allow the use of the Pythagorean theorem to simplify the calculations.