Introduction:
To prove that the altitudes BE and CF of an isosceles triangle ABC, drawn to equal sides AC and AB respectively, are equal, we can use the property of isosceles triangles and the concept of right triangles.

Proof:
Step 1: Identify the given information
- Triangle ABC is an isosceles triangle.
- Altitudes BE and CF are drawn from the vertices B and C to the equal sides AC and AB respectively.

Step 2: Understand the properties of an isosceles triangle
- In an isosceles triangle, the two equal sides are opposite the two equal angles.
- The altitude drawn from the vertex angle of an isosceles triangle bisects the base.

Step 3: Establish the equality of angles
- Since triangle ABC is isosceles, angle BAC = angle ABC (opposite sides are equal).
- Similarly, angle CBA = angle CAB.

Step 4: Establish the equality of altitudes
- Let's consider triangle ABE and triangle ACF.
- In triangle ABE, angle ABE = angle AEB (as the base angles of an isosceles triangle are equal).
- Similarly, in triangle ACF, angle ACF = angle AFC.

Step 5: Establish the congruence of right triangles
- Since angle ABE = angle ACF and angle AEB = angle AFC, we have angle ABE = angle ACF and angle AEB = angle AFC.
- Therefore, triangle ABE is congruent to triangle ACF by angle-angle-side congruence.

Step 6: Conclude the equality of altitudes
- Since triangle ABE is congruent to triangle ACF, their corresponding parts are also congruent.
- Therefore, BE = CF.

Conclusion:
By using the properties of isosceles triangles and proving the congruence of triangles, we have shown that the altitudes BE and CF of isosceles triangle ABC, drawn to equal sides AC and AB respectively, are equal.

BE=CF (given)AC=AB (given)angle E=angle F (90degree)