Question

ABCD is a rectangle with diagonals AC and DB.

Column A	Column B
\(r + u + v\)	\(r + u + w\)

Hint:

Do not assume a geometric figure has special properties unless stated. A rectangle is not necessarily a square. If the relationship changes when you consider a special case (like a square) versus a general case (like a non-square rectangle), the answer is (D).

Answer 1

Step 1: Understanding the Concept:
This problem tests knowledge of the properties of a rectangle, specifically its angles and diagonals. We need to compare two sums of angles.
Step 2: Key Formula or Approach:
Properties of a rectangle:
1. Opposite sides are parallel (AB \(||\) DC, AD \(||\) BC).
2. All corner angles are 90\(^\circ\).
3. Diagonals are equal in length and bisect each other.
Alternate interior angles formed by a transversal intersecting two parallel lines are equal.
Step 3: Detailed Explanation:
The comparison between Column A (\(r+u+v\)) and Column B (\(r+u+w\)) can be simplified by subtracting the common terms \(r+u\) from both columns. The problem then reduces to comparing \(v\) and \(w\).
Let's identify the angles from the diagram:
\(v\) is the measure of angle ABD.
\(w\) is the measure of angle CBD.
In rectangle ABCD, the angle at vertex B, \(\angle ABC\), is 90\(^\circ\). From the diagram, we can see that \(\angle ABC = \angle ABD + \angle CBD = v + w\). So, \(v+w = 90^\circ\).
Now, let's consider the relationship between \(v\) and \(w\).
Consider the right-angled triangle \(\triangle ABC\). We have \(\tan(u) = \frac{BC}{AB}\).
Consider the right-angled triangle \(\triangle ABD\). We have \(\tan(v) = \frac{AD}{AB}\).
Since ABCD is a rectangle, \(AD = BC\). Therefore, \(\tan(v) = \frac{BC}{AB}\). This means \(\tan(u) = \tan(v)\), so \(u=v\).
Now, let's relate \(v\) and \(w\). The diagonal DB divides the corner angle \(\angle ABC\) into two angles, \(v\) and \(w\). The diagonal bisects the corner angle only if the rectangle is a square (i.e., if adjacent sides are equal, \(AB=BC\)).
Case 1: The rectangle is a square.
If \(AB = BC\), then in \(\triangle ABC\), it is an isosceles right triangle, and the diagonal AC would bisect \(\angle B\). Similarly, diagonal DB would bisect \(\angle B\). Thus, \(v = w = 45^\circ\). In this case, Column A and Column B are equal.
Case 2: The rectangle is not a square.
Suppose \(AB>BC\). Consider the right-angled triangle \(\triangle BCD\). We have \(\tan(w) = \frac{DC}{BC}\). Since \(DC=AB\), \(\tan(w) = \frac{AB}{BC}\).
We are comparing \(v\) and \(w\). Let's compare \(\tan(v)\) and \(\tan(w)\).
\(\tan(v) = \frac{BC}{AB}\) and \(\tan(w) = \frac{AB}{BC}\).
Since we assumed \(AB>BC\), then \(\frac{AB}{BC}>1\) and \(\frac{BC}{AB}<1\).
So, \(\tan(w)>\tan(v)\). Since tangent is an increasing function for acute angles, this implies \(w>v\). In this case, Column B is greater.
If we assumed \(BC>AB\), we would find \(v>w\), making Column A greater.
Step 4: Final Answer:
Since the relationship between \(v\) and \(w\) depends on the side lengths of the rectangle (whether it is a square or an elongated rectangle), and this information is not given, we cannot determine the relationship between the columns.