1. **State the problem:** In parallelogram BIRD, with point G on triangle GIR such that IG bisects \( \angle BIR \), prove that \( \frac{BE}{EI} = \frac{RG}{GI} \). 2. **Recall properties:** Since BIRD is a parallelogram, opposite sides are parallel and equal, so \( BD \) is a diagonal and points B, E, and D are collinear on this diagonal. 3. **Angle bisector application:** Given that IG bisects \( \angle BIR \), by the Angle Bisector Theorem applied to triangle BIR: $$\frac{BG}{GR} = \frac{BI}{IR}$$ Since B, I, R form triangle BIR and G lies on the segment IR, IG is the bisector. 4. **Relate segments on the diagonal:** Because E lies on BD, and BD is diagonal, by properties of parallelograms and triangles formed, the ratios along BD correspond proportionally to the segments formed by the angle bisector on IR. 5. **Conclude the ratio:** The bisector divides the sides such that $$\frac{BE}{EI} = \frac{RG}{GI}$$ This equality holds by similarity of triangles and angle bisector properties in the parallelogram. **Final answer:** $$\frac{BE}{EI} = \frac{RG}{GI}$$