ABCD is a square. X is the mid-point of AB and Y is the mid-point of BC. Consider the following statements: 1. Triangles ADX and BAY are congruent. 2. ∠DXA = ∠AYB 3. DX is inclined at an angle 60° with AY. 4. DX is not perpendicular to AY. Which of the following statements are correct?
1 and 2 only
Analyzing Statements about Square ABCD and Midpoints
We are given a square ABCD. X is the mid-point of side AB and Y is the mid-point of side BC. We need to analyze four statements regarding triangles ADX and BAY and line segments DX and AY.
Evaluating Statement 1: Triangles ADX and BAY are congruent
Let the side length of the square ABCD be 's'.
- AD = AB = BC = CD = s (Sides of a square are equal).
- X is the midpoint of AB, so AX = XB = s/2.
- Y is the midpoint of BC, so BY = YC = s/2.
Consider triangles ADX and BAY:
- AD = BA (Sides of the square ABCD).
- \(\angle\)DAX = \(\angle\)ABY (Angles of the square ABCD are 90\(\degree\)).
- AX = BY (Both are equal to s/2, half the side length of the square).
By the SAS (Side-Angle-Side) congruence criterion, \(\triangle\)ADX is congruent to \(\triangle\)BAY (\(\triangle\)ADX \(\cong\) \(\triangle\)BAY).
Therefore, Statement 1 is correct.
Evaluating Statement 2: \(\angle\)DXA = \(\angle\)AYB
Since we have established that \(\triangle\)ADX \(\cong\) \(\triangle\)BAY from Statement 1, their corresponding parts are equal (CPCTC - Corresponding Parts of Congruent Triangles).
The angles \(\angle\)DXA and \(\angle\)AYB are corresponding angles in the congruent triangles \(\triangle\)ADX and \(\triangle\)BAY, respectively.
Therefore, \(\angle\)DXA = \(\angle\)AYB.
Statement 2 is correct.
Evaluating Statements 3 and 4: Angle between DX and AY
To analyze the angle between DX and AY, let's use coordinate geometry. Assume vertex D of the square is at the origin (0,0) and the side length is 'a'.
- D = (0,0)
- A = (0,a)
- B = (a,a)
- C = (a,0)
Since X is the midpoint of AB, the coordinates of X are \(\left(\frac{0+a}{2}, \frac{a+a}{2}\right) = \left(\frac{a}{2}, a\right)\).
Since Y is the midpoint of BC, the coordinates of Y are \(\left(\frac{a+a}{2}, \frac{a+0}{2}\right) = \left(a, \frac{a}{2}\right)\).
Now let's find the slopes of the line segments DX and AY.
Slope of DX (\(m_{DX}\)): Using points D(0,0) and X(\(a/2\), a))
\(m_{DX} = \frac{a - 0}{a/2 - 0} = \frac{a}{a/2} = 2\)
Slope of AY (\(m_{AY}\)): Using points A(0,a) and Y(\(a\), a/2))
\(m_{AY} = \frac{a/2 - a}{a - 0} = \frac{-a/2}{a} = -\frac{1}{2}\)
To find the angle between two lines with slopes \(m_1\) and \(m_2\), we can use the formula: \(\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|\). Alternatively, we can check for perpendicularity.
Let's calculate the product of the slopes \(m_{DX} \times m_{AY}\):
\(m_{DX} \times m_{AY} = 2 \times \left(-\frac{1}{2}\right) = -1\)
Since the product of the slopes of DX and AY is -1, the lines DX and AY are perpendicular to each other. The angle between perpendicular lines is 90\(\degree\).
- Statement 3 says DX is inclined at an angle 60\(\degree\) with AY. This is incorrect, as the angle is 90\(\degree\).
- Statement 4 says DX is not perpendicular to AY. This is incorrect, as DX is perpendicular to AY (angle is 90\(\degree\)).
Therefore, Statement 3 and Statement 4 are incorrect.
Summary of Statement Analysis
| Statement | Analysis | Correct/Incorrect |
|---|---|---|
| 1. Triangles ADX and BAY are congruent. | SAS congruence (\(\triangle\)ADX \(\cong\) \(\triangle\)BAY) | Correct |
| 2. \(\angle\)DXA = \(\angle\)AYB | Corresponding angles of congruent triangles | Correct |
| 3. DX is inclined at an angle 60\(\degree\) with AY. | Angle is 90\(\degree\) (perpendicular) | Incorrect |
| 4. DX is not perpendicular to AY. | DX is perpendicular to AY (slope product is -1) | Incorrect |
Based on the analysis, only Statement 1 and Statement 2 are correct.
Revision Table: Square Geometry Concepts
| Concept | Description |
|---|---|
| Square Properties | All four sides are equal, all four angles are 90\(\degree\), diagonals are equal and bisect each other at 90\(\degree\). |
| Midpoint | A point that divides a line segment into two equal parts. |
| Congruent Triangles | Triangles that have the same size and shape. Corresponding sides and corresponding angles are equal. Criteria include SSS, SAS, ASA, AAS, RHS. |
| Slope of a Line | The measure of the steepness of a line, calculated as the change in y divided by the change in x (\(\frac{y_2 - y_1}{x_2 - x_1}\)). |
| Perpendicular Lines | Two lines are perpendicular if they intersect at a 90\(\degree\) angle. In coordinate geometry, the product of their slopes is -1 (unless one line is vertical). |
Additional Information: Properties of Geometric Figures
Understanding the basic properties of geometric figures like squares and triangles is fundamental in solving geometry problems. Midpoints introduce specific length relationships that are often useful in congruence proofs or coordinate geometry calculations.
- Congruence: When triangles are congruent, it's a powerful tool because it implies the equality of all corresponding sides and angles. This is frequently used to prove other relationships in a diagram.
- Coordinate Geometry: Placing figures on a coordinate plane allows us to use algebraic methods (like calculating slopes, distances, and midpoints) to verify geometric properties and relationships, such as perpendicularity or parallelism.
- Slopes and Angles: The slope of a line is directly related to the angle it makes with the x-axis. The relationship between the slopes of two lines (parallel, perpendicular) tells us about the angle between them. For perpendicular lines (not vertical), the product of slopes is always -1.
In this problem involving the square ABCD and midpoints X and Y, applying both congruence principles for the triangles ADX and BAY and coordinate geometry to analyze the lines DX and AY helped confirm the correctness of the statements.
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