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Turkish Mathematical Olympiad

Geometry Difficulty 5.5 AIME, harder Prove it Turkey

Let $E$ and $F$ be two points on the side $CD$ of a convex quadrilateral $ABCD$ satisfying $0 < DE = FC < CD$ . Let $K$ be the second point of intersection of the circumcircles of the triangles $ADE$ and $ACF$ , and let $L$ be the second point intersection of the circumcircles of the triangles $BDE$ and $BCF$ . Show that the points $A$ , $B$ , $K$ , $L$ lie on a circle.

Solution

Let $\{M\} = DC \cap AK$ and $\{N\} = DC \cap BL$ . Considering the powers of the point $M$ with respect to the circles $ADE$ and $AFC$ we obtain
$ME \cdot MD = MK \cdot MA = MF \cdot MC.$
Since $DE = FC$ , we conclude that $M$ is the midpoint of the line segment $CD$ . Similarly,
$NE \cdot ND = NL \cdot NB = NF \cdot NC,$
and $N$ is the midpoint of the line segment $CD$ . Therefore, $M = N$ .

Now the equalities above give
$MK \cdot MA = ME \cdot MD = NE \cdot ND = NL \cdot NB = ML \cdot MB$
which implies that $K, A, L$ and $B$ lie on a circle.

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