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Geometry Difficulty 6.0 National Olympiad Prove it Slovenia

Let $ABCDE$ be a cyclic pentagon in which $|CD| = |DE|$ . Let the diagonals $AD$ and $BE$ intersect at point $K$ , and let the diagonals $AC$ and $BD$ intersect at point $L$ . Prove that the lines $KL$ and $EC$ are parallel.

Solution

According to the inscribed angle theorem, we have $\angle CED = \angle CAD$ and $\angle DCE = \angle DBE$ . Because $ECD$ is an isosceles triangle with the top angle at $D$ , we have $\angle CED = \angle DCE$ . Due to the collinearity of the points $A, L, C$ and the collinearity of the points $A, K, D$ we also have $\angle LAK = \angle LBK$ , hence points $A, B, L, K$ are concyclic. From this we get $\angle KLA = \angle KBA = \angle EBA = \angle ECA$ . The lines $KL$ and $EC$ are thus parallel.

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