All angles of a cyclic pentagon $ABCDE$ are obtuse. The sidelines $AB$ and $CD$ meet at point $E_1$, the sidelines $BC$ and $DE$ meet at point $A_1$. The tangent at $B$ to the circumcircle of the triangle $BE_1C$ meets the circumcircle $\omega$ of the pentagon for the second time at point $B_1$. The tangent at $D$ to the circumcircle of the triangle $DA_1C$ meets $\omega$ for the second time at point $D_1$. Prove that $B_1D_1 // AE$

Inside the convex pentagon $ABCDE$, a point $O$ was chosen, and it turned out that all five triangles $AOB$, $BOC$, $COD$, $DOE$ and $EOA$ are congrunet to each other. Prove that these triangles are isosceles or right-angled.

Let $ABCDE$ be a regular pentagon, and $F$ some point on the arc $AB$ of the circumcircle of $ABCDE$. Show that \[ \frac{FD}{FE + FC} = \frac{FB + FA}{FD} = \frac{-1 + \sqrt{5}}{2}, \] and that $FD + FB + FA = FE + FC$.

The pentagon $ABCDE$ is inscribed in the circle $\omega$. Let $T$ be the intersection point of the diagonals $BE$ and $AD$. A line is drawn through the point $T$ parallel to $CD$, which intersects $AB$ and $CE$ at points $X$ and $Y$, respectively. Prove that the circumscribed circle of the triangle $AXY$ is tangent to $\omega$.

Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$, $\angle{EAB}=\angle{BCD}$, and $\angle{EDC}=\angle{CBA}$. Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.

Call a convex polyhedron a [i]footballoid [/i] if it has the following properties. (1) Any face is either a regular pentagon or a regular hexagon. (2) All neighbours of a pentagonal face are hexagonal (a [i]neighbour [/i] of a face is a face that has a common edge with it). Find all possibilities for the number of pentagonal and hexagonal faces of a footballoid.

Given a convex pentagon $ABCDE$ with $\angle ABC= \angle ADE$ and $\angle AEC= \angle ADB$ . Prove that $\angle BAC = \angle DAE$ .

The pentagon $ABCDE$ is inscribed in the circle. Line segments $AC$ and $BD$ intersect at point $K$. Line segment $CE$ touches the circumcircle of triangle $ABK$ at point $N$. Find the angle $CNK$ if $\angle ECD = 40^o.$

An approximate construction of a regular pentagon goes as follows. Inscribe an arbitrary convex pentagon $P_1P_2P_3P_4P_5$ in a circle. Now choose an arror bound $\epsilon > 0$ and apply the following procedure. (a) Denote $P_0 = P_5$ and $P_6 = P_1$ and construct the midpoint $Q_i$ of the circular arc $P_{i-1}P_{i+1}$ containing $P_i$. (b) Rename the vertices $Q_1,...,Q_5$ as $P_1,...,P_5$. (c) Repeat this procedure until the difference between the lengths of the longest and the shortest among the arcs $P_iP_{i+1}$ is less than $\epsilon$. Prove this procedure must end in a finite time for any choice of $\epsilon$ and the points $P_i$.

A pentagon $ABCDE$ is circumscribed about a circle. The angles at the vertices $A{}$, $C{}$ and $E{}$ of the pentagon are equal to $100^\circ$. Find the measure of the angle $\angle ACE$.

Each diagonal of a convex pentagon is parallel to one side of the pentagon. Prove that the ratio of the length of a diagonal to that of its corresponding side is the same for all five diagonals, and compute this ratio.

$(NET 5)$ The bisectors of the exterior angles of a pentagon $B_1B_2B_3B_4B_5$ form another pentagon $A_1A_2A_3A_4A_5.$ Construct $B_1B_2B_3B_4B_5$ from the given pentagon $A_1A_2A_3A_4A_5.$

Let $ ABCDE$ be a convex pentagon such that \[ \angle BAC \equal{} \angle CAD \equal{} \angle DAE\qquad \text{and}\qquad \angle ABC \equal{} \angle ACD \equal{} \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$. [i]Proposed by Zuming Feng, USA[/i]

A rectangle of paper of $3$ cm by $9$ cm is folded along a straight line, making two opposite vertices coincide. In this way a pentagon is formed. Calculate it's area.

Let $ABCDE$ be a convex pentagon such that quadrilateral $ABDE$ is a parallelogram and quadrilateral $BCDE$ is inscribed in a circle. The circle with center $C$ and radius $CD$ intersects the line $BD, DE$ at points $F, G(\neq D)$, and points $A, F, G$ is on line l. Let $H$ be the intersection point of line $l$ and segment $BC$. Consider the set of circle $\Omega$ satisfying the following condition. Circle $\Omega$ passes through $A, H$ and intersects the sides $AB, AE$ at point other than $A$. Let $P, Q(\neq A)$ be the intersection point of circle $\Omega$ and sides $AB, AE$. Prove that $AP+AQ$ is constant.

The convex $n$-gon ($n\ge 5$) is cut along all its diagonals. Prove that there are at least a pair of parts with the different areas.

Let $ABCDE$ be a pentagon such that $\angle DCB < 90^{\circ} < \angle EDC$. The circle with diameter $BD$ intersects the line $BC$ again at $F$, and the circle with diameter $DE$ intersects the line $CE$ again at $G$. Prove that the second intersection ($\neq D$) of the circumcircle of $\triangle DFG$ and the circle with diameter $AD$ lies on $AC$. Proposed by [i]Petar Filipovski[/i]

A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$ . The line $BC$ intersects the circle $c$ for second time at point $F$. The point $G$ on the circle $c$ is chosen such that $| F B | = | FG |$ and $B \ne G$. Prove that the lines $AB, EF$ and $DG$ intersect at one point.

A pentagon $ABCDE$ is inscribed in a circle $O$, and satises $AB = BC , AE = DE$. The circle that is tangent to $DE$ at $E$ and passing $A$ hits $EC$ at $F$ and $BF$ at $G (\ne F)$. Let $DG\cap O = H (\ne D)$. Prove that the tangent to $O$ at $E$ is perpendicular to $HA$.

The convex pentagon $ABCDE$ was drawn in the plane. $A_1$ was symmetric to $A$ with respect to $B$. $B_1$ was symmetric to $B$ with respect to $C$. $C_1$ was symmetric to $C$ with respect to $D$. $D_1$ was symmetric to $D$ with respect to $E$. $E_1$ was symmetric to $E$ with respect to $A$. How is it possible to restore the initial pentagon with the compasses and ruler, knowing $A_1,B_1,C_1,D_1,E_1$ points?

Let $ ABCDE$ be a convex pentagon such that $ AB\plus{}CD\equal{}BC\plus{}DE$ and $ k$ a circle with center on side $ AE$ that touches the sides $ AB$, $ BC$, $ CD$ and $ DE$ at points $ P$, $ Q$, $ R$ and $ S$ (different from vertices of the pentagon) respectively. Prove that lines $ PS$ and $ AE$ are parallel.

On the side $DE$ and on the diagonal $BE$ of the regular pentagon $ABCDE$ we consider the squares $DEFG$ and $BEHI$. [list=a] [*] Prove that $A,I,$ and $G$ are collinear. [*] Prove that on this line lies also the centre $O$ of the square $BDJK$. [/list]

Is there a convex pentagon in which each diagonal is equal to a side?

Given a convex pentagon $ABCDE$ with $AB=BC, CD=DE, \angle ABC=150^o, \angle CDE=30^o, BD=2$. Find the area of $ABCDE$. (I.Voronovich)

There are $k$ points in the interior of a pentagon. Together with the vertices of the pentagon they form a $(k + 5)$-element set $M$. The area of the pentagon is defined by connecting lines between the points of $M$ into sub-areas in such a way that it is divided into sub-areas in such a way that no sub-areas have a point on their interior of $M$ and contains exactly three points of $M$ on the boundary of each part. None of the connecting lines has a point in common with any other connecting line or pentagon side, which does not belong to $M$. With such a decomposition of the pentagon, there can be an even number of connecting lines (including the pentagon sides) go out? The answer has to be justified.