ABC is a triangle in which altitude BE and CF drawn on AC and AB respectively are equal. Prove that `DeltaBEC~=DeltaCFB`

Answer

To prove that triangles \( \triangle BEC \) and \( \triangle CFB \) are congruent, we will use the criteria for congruence of triangles, specifically the RHS (Right angle-Hypotenuse-Side) criterion.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- We have triangle \( ABC \).
- Altitudes \( BE \) and \( CF \) are drawn from points \( B \) and \( C \) to sides \( AC \) and \( AB \) respectively.
- It is given that \( BE = CF \).

**Hint**: Write down all the given information clearly to visualize the problem.

2. **Draw the Triangle**:
- Sketch triangle \( ABC \) with points \( B \) and \( C \) at the vertices, and draw the altitudes \( BE \) and \( CF \) to sides \( AC \) and \( AB \).

**Hint**: Visualizing the triangle helps in understanding the relationships between the segments.

3. **Identify Right Angles**:
- Since \( BE \) and \( CF \) are altitudes, we know that:
\[ \angle BEC = 90^\circ \quad \text{and} \quad \angle CFB = 90^\circ \]

**Hint**: Remember that altitudes create right angles with the base.

4. **State the Equal Lengths**:
- From the problem statement, we know:
\[ BE = CF \]

**Hint**: Identify and write down all equal lengths as they will be crucial for proving congruence.

5. **Identify the Common Side**:
- The side \( BC \) is common to both triangles \( \triangle BEC \) and \( \triangle CFB \):
\[ BC = BC \quad (\text{common side}) \]

**Hint**: Look for sides that are shared between the two triangles.

6. **Apply the RHS Congruence Criterion**:
- We have established:
- \( \angle BEC = 90^\circ \)
- \( \angle CFB = 90^\circ \)
- \( BE = CF \)
- \( BC = BC \) (common side)
- Therefore, by the RHS criterion (Right angle, Hypotenuse, Side), we can conclude:
\[ \triangle BEC \cong \triangle CFB \]

**Hint**: Recall the RHS criterion: if two triangles have one right angle, the hypotenuse is equal, and one side is equal, they are congruent.

7. **Conclusion**:
- Hence, we have proved that \( \triangle BEC \) is congruent to \( \triangle CFB \).

**Hint**: Always summarize your findings to reinforce the conclusion.