The diagonals of a convex quadrilateral $ABCD$ meet at point $L$. The orthocenter $H$ of the triangle $LAB$ and the circumcenters $O_1, O_2$, and $O_3$ of the triangles $LBC, LCD$, and $LDA$ were marked. Then the whole configuration except for points $H, O_1, O_2$, and $O_3$ was erased. Restore it using a compass and a ruler.

$(CZS 5)$ 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 such that the triangle $ABD$ is equilateral and the triangle $BCD$ is isosceles, with $\angle C = 90^o$. If $E$ is the midpoint of the side $AD$, determine the measure of the angle $\angle CED$.

A convex quadrilateral $ABCD$ Is placed on the Cartesian plane. Its vertices $A$ and $D$ belong to the negative branch of the graph of the hyperbola $y= 1/x$, the vertices $B$ and $C$ belong to the positive branch of the graph and point $B$ lies at the left of $C$, the segment $AC$ passes through the origin $(0,0)$. Prove that $\angle BAD = \angle BCD$. (I, Voronovich)

The diagonals of a convex quadrilateral dissect it into four similar triangles. Prove that this quadrilateral can also be dissected into two congruent triangles.

Let $S$ be a convex quadrilateral $ABCD$ and $O$ a point inside it. The feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ are $A_1, B_1, C_1, D_1$ respectively. The feet of the perpendiculars from $O$ to the sides of $S_i$, the quadrilateral $A_iB_iC_iD_i$, are $A_{i+1}B_{i+1}C_{i+1}D_{i+1}$, where $i = 1, 2, 3.$ Prove that $S_4$ is similar to S.

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.

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

Prove that there exists a unique point $M$ on the side $AD$ of a convex quadrilateral $ABCD$ such that \[\sqrt{S_{ABM}}+\sqrt{S_{CDM}} = \sqrt{S_{ABCD}}\] if and only if $AB\parallel CD$.

Prove that in every convex quadrilateral of area $1$, the sum of the lengths of the sides and diagonals is not smaller than $2(2+\sqrt2)$.

Let $S$ be a convex quadrilateral $ABCD$ and $O$ a point inside it. The feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ are $A_1, B_1, C_1, D_1$ respectively. The feet of the perpendiculars from $O$ to the sides of $S_i$, the quadrilateral $A_iB_iC_iD_i$, are $A_{i+1}B_{i+1}C_{i+1}D_{i+1}$, where $i = 1, 2, 3.$ Prove that $S_4$ is similar to S.

Find all nonconvex quadrilaterals in which the sum of the distances to the lines containing the sides is the same for any interior point. Try to generalize the result in the case of an arbitrary non-convex polygon, polyhedron.

Given a convex quadrilateral $ABCD$. Let $O$ be the intersection point of diagonals $AC$ and $BD$ and let $I , K , H$ be feet of perpendiculars from $B , O , C$ to $AD$, respectively. Prove that $AD \times BI \times CH \le AC \times BD \times OK$.

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$.

In a convex quadrilateral it is known $ABCD$ that $\angle ADB + \angle ACB = \angle CAB + \angle DBA = 30^{\circ}$ and $AD = BC$. Prove that from the lengths $DB$, $CA$ and $DC$, you can make a right triangle.

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 $MNPQ$?

Let $S$ be a convex quadrilateral $ABCD$ and $O$ a point inside it. The feet of the perpendiculars from $O$ to $AB, BC, CD, DA$ are $A_1, B_1, C_1, D_1$ respectively. The feet of the perpendiculars from $O$ to the sides of $S_i$, the quadrilateral $A_iB_iC_iD_i$, are $A_{i+1}B_{i+1}C_{i+1}D_{i+1}$, where $i = 1, 2, 3.$ Prove that $S_4$ is similar to S.

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 with $AB=CD$. From a random point $P$ of it's diagonal $BD$, we draw a line parallel to $AB$ that intersects $AD$ at point $M$ and a line parallel to $CD$ that intersects $BC$ at point $N$. Prove that: a) The sum $PM+PN$ is constant, independent of the position of $P$ on the diagonal $BD$. b) $MN\le BD$. When the equality holds?

Let \(ABCD\) be a convex quadrilateral in the plane and let \(O_{A}, O_{B}, O_{C}\) and \(O_{D}\) be the circumcenters of the triangles \(BCD, CDA, DAB\) and \(ABC\), respectively. Suppose these four circumcenters are distinct points. Prove that these points are not on a same circle.

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.$

A subset $\bf{S}$ of the plane is called [i]convex[/i] if given any two points $x$ and $y$ in $\bf{S}$, the line segment joining $x$ and $y$ is contained in $\bf{S}$. A quadrilateral is called [i]convex[/i] if the region enclosed by the edges of the quadrilateral is a convex set. Show that given a convex quadrilateral $Q$ of area $1$, there is a rectangle $R$ of area $2$ such that $Q$ can be drawn inside $R$.

A convex flat quadrilateral is given. Prove that for the ratio $q$ of the largest to the smallest of all distances, for any two vertices: $q \ge \sqrt2$. [hide=original wording]Gegeben sei ein konvexes ebenes Viereck. Es ist zu beweisen, dass fur den Quotienten q aus dem großten und dem kleinsten aller Abstande zweier beliebiger Eckpunkte voneinander stets gilt: q >= \sqrt2.[/hide]

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.

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 $.