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Geometry Difficulty 6.7 National Olympiad Prove it Slovenia

Let $D$ and $E$ be such points on the sides $BC$ and $AC$ of a triangle $ABC$ , respectively, that the points $A$ , $B$ , $D$ and $E$ lie on the same circle. Let $L$ denote the center of the inscribed circle of the triangle $BCE$ , and let $G$ denote the point of tangency of this inscribed circle with the side $EC$ . Let $K$ denote the center of the inscribed circle of the triangle $DCA$ , and let $F$ denote the point of tangency of this inscribed circle with the side $DC$ . Let $N$ be the intersection point of the lines $EL$ and $DK$ , and let $M$ be the intersection point of the lines $KF$ and $LG$ . Prove that the points $A$ , $B$ , $D$ and $N$ lie on the same circle and that $KMLN$ is a deltoid.

Solution

Let $K$ be a circle containing points $A$ , $B$ , $D$ and $E$ . Because $EL$ is the bisector of the angle $\angle BEC$ , it bisects the arc $\text{AB}$ that contains $E$ . Similarly, the line $DK$ bisects the arc $\text{AB}$ that contains $D$ . Because $E$ and $D$ lie on the same side of the line $\text{AB}$ , the lines $EL$ and $DK$ bisect the same arc $\text{AB}$ and their point of intersection lies on the circle $K$ (it is the point $N$ ). Hence, the points $A$ , $B$ , $D$ , $N$ are concyclic.

Because the points $A$ , $B$ , $D$ , $E$ are concyclic, the angles $\angle CDA$ and $\angle BEC$ are equal. If we thus denote $\alpha = \angle CDK$ , we have $\angle CDK = \angle KDA = \angle BEL = \angle LEC = \alpha$ . Since the points $K$ , $L$ and $C$ are colinear, the following holds:
$\begin{align*} \angle NLK &= 180^\circ - CLE = \angle ECL + \alpha = \\ &= \angle KCD + \alpha = 180^\circ - \angle DKC = \\ &= \angle LKN. \end{align*}$
We can show similarly that $\angle MKL = \angle KLM$ . The quadrilateral $KMLN$ is thus a deltoid.

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