In a triangle $ABC$, let $D,\ E$ be the intersection points of the bisectors of $\angle{B},\ \angle{C}$ and these opposite sides respectively and let denote $I$ as the intersection point of the bisectors. If $ID=IE$, then determine the shape of $\triangle{ABC}.$

1956 Tokyo Institute of Technology entrance exam
1. **Draw the perpendiculars and identify congruent triangles:**
   - Draw the perpendiculars \( IY \) and \( IZ \) from \( I \) to \( AC \) and \( AB \) respectively.
   - Given \( ID = IE \), \( IY = IZ \), and \(\angle IYD = 90^\circ = \angle IZE \), triangles \( IYD \) and \( IZE \) are congruent by the RHS (Right angle-Hypotenuse-Side) congruence criterion.

2. **Analyze the congruence implications:**
   - Since \( IYD \cong IZE \), it follows that \( \angle IYD = \angle IZE \) and \( IY = IZ \).
   - This implies that \( I \) is equidistant from \( AB \) and \( AC \), suggesting symmetry in the triangle.

3. **Case I: Both \(\angle ADI\) and \(\angle AEI\) are acute, right, or obtuse:**
   - If both angles are of the same type (acute, right, or obtuse), then \( AY = AZ \).
   - This symmetry implies that \( AB = AC \), making \( \triangle ABC \) isosceles with \( AB = AC \).

4. **Case II: One of \(\angle ADI\) and \(\angle AEI\) is acute, and the other is obtuse:**
   - Assume without loss of generality that \(\angle ADI < 90^\circ\) and \(\angle AEI > 90^\circ\).
   - Notice that quadrilateral \( ADIE \) is cyclic because opposite angles sum to \( 180^\circ \).
   - Therefore, \(\angle BDA + \angle CEA = 180^\circ\).

5. **Calculate the angles:**
   - Given \(\angle BDA = \frac{\angle B}{2} + \angle C\) and \(\angle CEA = \angle B + \frac{\angle C}{2}\), we have:
     \[
     \left(\frac{\angle B}{2} + \angle C\right) + \left(\angle B + \frac{\angle C}{2}\right) = 180^\circ
     \]
   - Simplifying, we get:
     \[
     \frac{\angle B}{2} + \angle C + \angle B + \frac{\angle C}{2} = 180^\circ
     \]
     \[
     \frac{3\angle B}{2} + \frac{3\angle C}{2} = 180^\circ
     \]
     \[
     \angle B + \angle C = 120^\circ
     \]
   - Since the sum of the angles in a triangle is \( 180^\circ \), we have:
     \[
     \angle A = 180^\circ - (\angle B + \angle C) = 180^\circ - 120^\circ = 60^\circ
     \]

Conclusion:
- In both cases, we find that \( \triangle ABC \) is either isosceles with \( AB = AC \) or has \( \angle A = 60^\circ \). However, since \( \angle A = 60^\circ \) implies that \( \triangle ABC \) is equilateral (as all angles in an equilateral triangle are \( 60^\circ \)), we conclude that \( \triangle ABC \) must be equilateral.

The final answer is \( \boxed{ \triangle ABC } \) is equilateral.