Grade 6 question using segment addition and subtration properties .On a straight line xy place in order the points A C B D with AB=CD and  AC=BD Let point I  be midpoint of BC and midpoint of AD.
  . Trace the straight line d perpendicular to xy at C let m be an arbitrary point of d . Is ma=mb. Draw graph and justify 

To solve this problem, let's break it down step by step using the properties of segment addition and subtraction.

### Given Information:
1. Points A, C, B, D are on a straight line in the order A, C, B, D.
2. \( AB = CD \)
3. \( AC = BD \)
4. Point I is the midpoint of BC and the midpoint of AD.
5. Line d is perpendicular to xy at point C, and point M is an arbitrary point on line d.

### Steps to Solve:

1. **Set Up the Points:**
   - Place the points on a coordinate system.
   - Let \( A(0, 0) \), \( C(a, 0) \), \( B(a + b, 0) \), and \( D(a + b + c, 0) \), where \( b \) is the length of segment AB and \( c \) is the length of segment CD.
   - Given \( AB = CD \), \( b = c \).

2. **Find Coordinates for Points:**
   - Since \( I \) is the midpoint of both segments:
     - \( I \) for BC: 
       \[
       I = \left(\frac{a + (a + b)}{2}, 0\right) = \left(a + \frac{b}{2}, 0\right)
       \]
     - \( I \) for AD:
       \[
       I = \left(\frac{0 + (a + b + c)}{2}, 0\right) = \left(\frac{a + b + c}{2}, 0\right)
       \]
     - Set \( b = c \) implies that \( a + \frac{b}{2} = \frac{a + 2b}{2} \).

3. **Draw the Perpendicular Line:**
   - At point \( C(a, 0) \), draw line \( d \) which is vertical, meaning \( M \) can be any point of the form \( (a, y) \) where \( y \) is not equal to 0.

4. **Show that \( MA = MB \):**
   - The distances from \( M \) to \( A \) and \( B \) can be calculated:
     - Distance \( MA \):
       \[
       MA = \sqrt{(a - 0)^2 + (y - 0)^2} = \sqrt{a^2 + y^2}
       \]
     - Distance \( MB \):
       \[
       MB = \sqrt{(a - (a + b))^2 + (y - 0)^2} = \sqrt{(-b)^2 + y^2} = \sqrt{b^2 + y^2}
       \]
   - Since \( AB = CD \) (or \( b = c \)), and with \( b = c \), \( MA = MB \).

### Conclusion:
Since both distances are equal for any arbitrary point \( M \) on line \( d \), we conclude that \( MA = MB \).

### Graph:
To visualize this, you can draw:
- A horizontal line with points A, C, B, and D marked at appropriate distances.
- A vertical line at point C.
- Points A, B, and M marked clearly.

### Justification:
The geometric properties of midpoints and the equal distances from a perpendicular point to two points on a straight line show that the distances \( MA \) and \( MB \) remain equal.

### Questions for Further Understanding:
1. How would the distances change if \( b \neq c \)?
2. What happens if point M lies on the x-axis instead of line d?
3. Can you identify similar properties in other geometric figures?
4. How would you prove the midpoint theorem using different shapes?
5. What role does symmetry play in this problem?

### Tip:
When solving geometric problems, always visualize them with a sketch to better understand relationships and properties.

Grade 6 question using segment addition and subtraction properties. On a straight line xy, place in order the points A, C, B, D with AB=CD and AC=BD. Let point I be the midpoint of BC and midpoint of AD. Trace the straight line d perpendicular to xy at C; let m be an arbitrary point of d. Is MA=MB? Draw a graph and justify.

Explore a Grade 6 geometry problem involving points on a line with properties of segment addition and subtraction. Learn how to determine if the distances MA and MB are equal when M is on a perpendicular line. This problem covers concepts of midpoints, distance formula, and perpendicular lines, with a detailed solution and graph.

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