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The problem described on the board refers to a right-angled triangle \( \triangle ABC \), with the following information:

- \( AC = 10 \, \text{cm} \)
- \( BC = 8 \, \text{cm} \)
- \( AB \), the hypotenuse of the right-angled triangle at \( B \), can be calculated as \( AB = \sqrt{AC^2 - BC^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 \, \text{cm} \).

Additionally:
- \( D \) is the midpoint of \( BC \), so \( BD = DC = \frac{8}{2} = 4 \, \text{cm} \).
- \( E \) is the midpoint of \( AC \), so \( AE = EC = \frac{10}{2} = 5 \, \text{cm} \).

It also states that the intersection of lines \( AD \) and \( BE \) is the point \( M \), with the condition that \( AD \perp BE \).

We are asked to:
- Draw line \( CM \) that cuts \( AB \) at a point \( F \).
- Find the perimeter of \( \triangle AFE \).

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### Step-by-Step Breakdown:

#### 1. **Calculate Key Lengths:**
   - **Midpoints:** We already have \( BD = DC = 4 \, \text{cm} \) and \( AE = EC = 5 \, \text{cm} \).
   - **\( AB \):** As calculated, \( AB = 6 \, \text{cm} \).

#### 2. **Understanding the Problem:**
   - The triangle \( AFE \) involves segments of \( AC \) and \( AB \), with the line \( CM \) cutting \( AB \) at \( F \).

#### 3. **Coordinates for Simplicity (Optional Step):**
   Assign coordinates for each vertex:
   - Let \( B = (0, 0) \), \( C = (8, 0) \), and \( A = (0, 10) \).
   - \( D \), the midpoint of \( BC \), is \( D = (4, 0) \).
   - \( E \), the midpoint of \( AC \), is \( E = (0, 5) \).
   - The intersection points and calculation for \( F \) could then be approached geometrically using these coordinates.

#### 4. **Perimeter of \( \triangle AFE \):**
   To find the perimeter of \( \triangle AFE \), we need to calculate the lengths of:
   - \( AF \)
   - \( FE \)
   - \( AE \)

#### 5. **Conclusion and Further Calculations:**
   To complete the solution, more information would be needed on how to locate the point \( F \) along line \( AB \). This might involve deeper exploration using the geometry of the triangle.

Would you like to proceed with more detailed calculations on specific parts of this problem, or focus on a particular step? Let me know if you need any clarification or have additional questions.

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**Relative questions for further exploration:**
1. How do we find the coordinates of point \( F \) geometrically using the midline theorem?
2. Can you explain how the property of midpoints can help in finding the perpendicular intersections?
3. What is the significance of the intersection point \( M \) in relation to the triangle's geometry?
4. How does Pythagoras' theorem simplify calculations in right-angled triangles like \( \triangle ABC \)?
5. Can the concept of similar triangles be applied to simplify parts of this problem?

**Tip:** Using symmetry and midpoints often simplifies the calculation of perimeters in geometric problems like this one.

ABC is a right-angled triangle at B in which: AC = 10 cm, BC = 8 cm. D and E are the midpoints of BC and AC respectively where AD ⊥ BE = 8m3. Draw CM to cut AB at F. Find the perimeter of ΔAFE.

This geometry problem involves finding the perimeter of triangle AFE using the midline and Pythagorean theorem in a right-angled triangle where AC = 10 cm and BC = 8 cm. Suitable for students in Grades 8-10.

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