Prove that an arbitrary convex quadrilateral can be divided into five polygons having symmetry axes. (N. Belukhov)

A convex quadrilateral $ABC$ is given. $A',B',C',D'$ are the orthocenters of triangles $BCD, CDA, DAB, ABC$ respectively. Prove that in the quadrilaterals $ABCP$ and $A'B'C'D'$, the corresponding diagonals share the intersection points in the same ratio.

In the convex quadrilateral $ABCD$, point $K$ is the midpoint of $AB$, point $L$ is the midpoint of $BC$, point $M$ is the midpoint of CD, and point $N$ is the midpoint of $DA$. Let $S$ be a point lying inside the quadrilateral $ABCD$ such that $KS = LS$ and $NS = MS$ .Prove that $\angle KSN = \angle MSL$.

In a convex quadrilateral $ABCD$ the angles $\angle A$ and $\angle C$ are equal and the bisector of $\angle B$ passes through the midpoint of the side $CD$. If it is known that $CD = 3AD$, calculate $\frac{AB}{BC}$.

Let $ABCD$ be a convex quadrilateral and draw regular triangles $ABM, CDP, BCN, ADQ$, the first two outward and the other two inward. Prove that $MN = AC$. What can be said about the quadrilateral $MNPQ$?

Given $n>4$ points in the plane, no three collinear. Prove that there are at least $\frac{(n-3)(n-4)}{2}$ convex quadrilaterals with vertices amongst the $n$ points.

$(CZS 5)$ A convex quadrilateral $ABCD$ with sides $AB = a, BC = b, CD = c, DA = d$ and angles $\alpha = \angle DAB, \beta = \angle ABC, \gamma = \angle BCD,$ and $\delta = \angle CDA$ is given. Let $s = \frac{a + b + c +d}{2}$ and $P$ be the area of the quadrilateral. Prove that $P^2 = (s - a)(s - b)(s - c)(s - d) - abcd \cos^2\frac{\alpha +\gamma}{2}$

Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.

Let $ABCD$ be a convex quadrilateral, and $P$ be a point in its interior. The projections of $P$ on the sides of the quadrilateral lie on a circle with center $O$. Show that $O$ lies on the line through the midpoints of $AC$ and $BD$.

Suppose two convex quadrangles are such that the sides of each of them lie on the perpendicular bisectors of the sides of the other one. Determine their angles,

Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX \equal{} \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX \equal{} \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB \equal{} 2\cdot\measuredangle ADX$.

The $ AB, BC $ and $ CD $ segments of the polygon $ ABCD $ have the same length and are tangent to a circle $ S $, centered on the point $ O $. Let $ P $ be the point of tangency of $ BC $ with $ S $, and let $ Q $ be the intersection point of lines $ AC $ and $ BD $. Show that the point $ Q $ is collinear with the points $ P $ and $ O $.

A convex quadrilateral has equal diagonals. An equilateral triangle is constructed on the outside of each side of the quadrilateral. The centers of the triangles on opposite sides are joined. Show that the two joining lines are perpendicular. [i]Alternative formulation.[/i] Given a convex quadrilateral $ ABCD$ with congruent diagonals $ AC \equal{} BD.$ Four regular triangles are errected externally on its sides. Prove that the segments joining the centroids of the triangles on the opposite sides are perpendicular to each other. [i]Original formulation:[/i] Let $ ABCD$ be a convex quadrilateral such that $ AC \equal{} BD.$ Equilateral triangles are constructed on the sides of the quadrilateral. Let $ O_1,O_2,O_3,O_4$ be the centers of the triangles constructed on $ AB,BC,CD,DA$ respectively. Show that $ O_1O_3$ is perpendicular to $ O_2O_4.$

Let $P$ be a point inside a convex quadrilateral $ABCD$. Points $K,L,M,N$ are given on the sides $AB,BC,CD,DA$ respectively such that $PKBL$ and $PMDN$ are parallelograms. Let $S,S_1$, and $S_2$ be the areas of $ABCD, PNAK$, and $PLCM$. Prove that $\sqrt{S}\ge \sqrt{S_1} +\sqrt{S_2}$.

Let $ABCD$ be a convex quadrilateral and $S$ an arbitrary point in its interior. Let also $E$ be the symmetric point of $S$ with respect to the midpoint $K$ of the side $AB$ and let $Z$ be the symmetric point of $S$ with respect to the midpoint $L$ of the side $CD$. Prove that $(AECZ) = (EBZD) = (ABCD)$.