Školjka

%V0 $ABCD$ is a quadrilateral with $BC$ parallel to $AD$. $M$ is the midpoint of $CD$, $P$ is the midpoint of $MA$ and $Q$ is the midpoint of $MB$. The lines $DP$ and $CQ$ meet at $N$. Prove that $N$ is inside the quadrilateral $ABCD$.

%V0 Let $A, B$ and $C$ be non-collinear points. Prove that there is a unique point $X$ in the plane of $ABC$ such that $$XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2.$$

%V0 Let $ABC$ be a trapezoid with parallel sides $AB > CD$. Points $K$ and $L$ lie on the line segments $AB$ and $CD$, respectively, so that $\frac {AK}{KB} = \frac {DL}{LC}$. Suppose that there are points $P$ and $Q$ on the line segment $KL$ satisfying $\angle{APB} = \angle{BCD}$ and $\angle{CQD} = \angle{ABC}$. Prove that the points $P$, $Q$, $B$ and $C$ are concylic.

%V0 Consider a convex pentagon $ABCDE$ such that $$\angle BAC = \angle CAD = \angle DAE\ \ \ ,\ \ \ \angle ABC = \angle ACD = \angle ADE$$ Let $P$ be the point of intersection of the lines $BD$ and $CE$. Prove that the line $AP$ passes through the midpoint of the side $CD$.

%V0 The diagonals of a trapezoid $ABCD$ intersect at point $P$. Point $Q$ lies between the parallel lines $BC$ and $AD$ such that $\angle AQD = \angle CQB$, and line $CD$ separates points $P$ and $Q$. Prove that $\angle BQP = \angle DAQ$. Author: unknown author, Ukraine

%V0 Let $ABCD$ be a convex quadrilateral and let $P$ and $Q$ be points in $ABCD$ such that $PQDA$ and $QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $E$ on the line segment $PQ$ such that $\angle PAE = \angle QDE$ and $\angle PBE = \angle QCE$. Show that the quadrilateral $ABCD$ is cyclic. Proposed by John Cuya, Peru