LI OM - II - Problem 4

Point $ I $ is the center of the circle inscribed in triangle $ ABC $, where $ AB \neq AC $. Lines $ BI $ and $ CI $ intersect sides $ AC $ and $ AB $ at points $ D $ and $ E $, respectively. Determine all possible measures of angle $ BAC $ for which the equality $ DI = EI $ can hold.
We will show that the only value taken by angle $ BAC $ is $ 60^\circ $. 
By the Law of Sines applied to triangles $ ADI $ and $ AEI $, we obtain $ \sin \measuredangle AEI = \sin \measuredangle ADI $. Hence,






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First, suppose that the equality $ \measuredangle AEI = \measuredangle ADI $ holds (Fig. 1). Then also $ \measuredangle AIE = \measuredangle AID $, which means that triangles $ AEI $ and $ ADI $ are congruent (angle-side-angle criterion). Therefore, $ AD = AE $. This proves that triangles $ ADB $ and $ AEC $ are also congruent (angle-side-angle criterion). Hence, we obtain $ AB = AC $, which contradicts the assumptions made in the problem statement.
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The remaining case to consider is when $ \measuredangle AEI + \measuredangle ADI = 180^\circ $ (Fig. 2). Then points $ A $, $ E $, $ I $, $ D $ lie on a single circle. Therefore,






From this, we obtain






which means $ \measuredangle BAC = 60^\circ $.
To complete the solution, it remains to show that there exists a triangle $ ABC $ in which $ AB \neq AC $, $ \measuredangle BAC = 60^\circ $, and $ DI = EI $. We will show more: in any triangle $ ABC $ with $ \measuredangle BAC = 60^\circ $, the equality $ DI = EI $ holds.
If $ \measuredangle BAC = 60^\circ $, then






Therefore, a circle can be circumscribed around quadrilateral $ AEID $. Since $ AI $ is the angle bisector of $ \angle EAD $, then $ DI = EI $.