Given that $\triangle ABC$ is an isosceles triangle, we know that $AB = AC$. Altitudes $BE$ and $CF$ are drawn from vertices $B$ and $C$ to the opposite equal sides $AC$ and $AB$, respectively.

To prove:

$BE = CF$, i.e., the altitudes are equal.

Steps:

1. Congruence of triangles: Consider $\triangle ABE$ and $\triangle ACF$.

   • $AB = AC$ (given, as $\triangle ABC$ is isosceles).
   • $\angle AEB = \angle AFC = 90^\circ$ (since both $BE$ and $CF$ are altitudes).
   • $AE = AF$ (as $E$ and $F$ are the feet of the altitudes on equal sides, and in an isosceles triangle, the perpendicular distances from the base to equal sides are equal).

   By RHS Congruence (Right-angle Hypotenuse-Side Criterion), we have: $\triangle ABE \cong \triangle ACF$

2. Corresponding parts of congruent triangles are equal (CPCT): Since $\triangle ABE \cong \triangle ACF$, we can conclude that: $BE = CF$ Thus, the altitudes are equal.

Conclusion:

The altitudes $BE$ and $CF$ in the isosceles triangle $ABC$ are equal.

---

Would you like to go over the details, or do you have any questions about this?

Here are 5 related questions to expand this topic:

1. What is the significance of the RHS congruence criterion in geometry?
2. How would the proof change if $\triangle ABC$ was not isosceles?
3. Can you prove the equality of altitudes using coordinate geometry?
4. How do the altitudes in an equilateral triangle compare to those in an isosceles triangle?
5. What are the properties of the centroid in an isosceles triangle?

Tip: When proving congruence, always check for at least one of the established congruence criteria (SSS, SAS, ASA, RHS) to ensure valid reasoning.