N.T.TUAN wrote: Given a triangle. A variable poin $ D$ is chosen on side $ BC$. Points $ K$ and $ L$ are the incenters of triangles $ ABD$ and $ ACD$, respectively. Prove that the second intersection point of the circumcircles of triangles $ BKD$ and $ CLD$ moves along on a fixed circle (while $ D$ moves along segment $ BC$). Invalid URL We have $ \angle BKD = 90+\frac{\angle BAD}{2}$ and $ \angle CLD = 90+\frac{\angle CAD}{2}$ Let $ E$ be the 2nd intersection point of the two circumcircles. Then $ \angle BED = 180-\angle BKD$ and $ \angle CED = 180-\angle CLD$ (*) $ \angle BEC =\angle BED+\angle CED$ $ = 360-(\angle BKD+\angle CLD)$ $ = 360-180-\frac{\angle BAD+\angle CAD}{2}$ $ = 180-\frac{\angle BAC}{2}$ So $ E$ belongs to the arc of a circle The center $ O$ of this circle belongs to the circumcircle of $ \triangle ABC$ and to the perpendicular bisector of $ BC$ And it is with $ A$ on the same side of $ BC$ because we want $ \angle BOC =\angle BAC$ $ \star$ Note Since $ BKD>90$ we get that the circumcenter of $ \triangle BKD$ and the point $ A$ is on different sides of $ BC$. We will say that the circumcenter is "below" of $ BC$ Similarly, the circumcenter of $ \triangle CDL$ is also below of $ BC$. Then their midpoint and consequently the point $ E$ are also below of $ BC$. So $ \angle BKD+\angle BED = 180$

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