Let $ABCD$ be a parallelogram and $l$ the line parallel to $AC$ which passes through $D$. Let $E$ and $F$ points on $l$ such that $DE=DF=DB$. Show that $EA,FC$ and $BD$ are concurrent.

In triangle $ABC$, point $I$ is the center of the inscribed circle. $AT$ is a segment tangent to the circle circumscribed around the triangle $BIC$ . On the ray $AB$ beyond the point$ B$ and on the ray $AC$ beyond the point $C$, we draw the segments $BD$ and $CE$, respectively, such that $BD = CE = AT$. Let the point $F$ be such that $ABFC$ is a parallelogram. Prove that points $D, E$ and $F$ lie on the same line. (Dmitry Prokopenko)

The diagonals of a parallelogram $ABCD$ meet at point $O$. The tangent to the circumcircle of triangle $BOC$ at $O$ meets ray $CB$ at point $F$. The circumcircle of triangle $FOD$ meets $BC$ for the second time at point $G$. Prove that $AG=AB$.

A point $P$ was chosen on the smaller arc $BC$ of the circumcircle of the acute-angled triangle $ABC$. Points $R$ and $S$ on the sides$ AB$ and $AC$ are respectively selected so that $CPRS$ is a parallelogram. Point $T$ on the arc $AC$ of the circumscribed circle of $\vartriangle ABC$ such that $BT \parallel CP$. Prove that $\angle TSC = \angle BAC$. (Anton Trygub)

A parallelogram $ABCD$ with $|AD| =|BD|$ has been given. A point $E$ lies on line segment $|BD|$ in such a way that $|AE| = |DE|$. The (extended) line $AE$ intersects line segment $BC$ in $F$. Line $DF$ is the angle bisector of angle $CDE$. Determine the size of angle $ABD$. [asy] unitsize (3 cm); pair A, B, C, D, E, F; D = (0,0); A = dir(250); B = dir(290); C = B + D - A; E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, D); F = extension(A, E, B, C); draw(A--B--C--D--cycle); draw(A--F--D--B); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, S); dot("$F$", F, SE); [/asy]

On the plane, there is drawn a parallelogram $P$ and a point $X$ outside of $P$. Using only an ungraded rule, determine the point $W$ that is symmetric to $X$ with respect to the center $O$ of $P$.

Let $ABCD$ be a parallelogram such that $|AB| > |BC|$. Let $O$ be a point on the line $CD$ such that $|OB| = |OD|$. Let $\omega$ be a circle with center $O$ and radius $|OC|$. If $T$ is the second intersection of $\omega$ and $CD$, prove that $AT, BO$ and $\omega$ are concurrent. [i]Proposed by Ivan Novak[/i]

Let $S$ be the set of points $A$ in the Cartesian plane such that the four points $A$, $(2, 3)$, $(-1, 0)$, and $(0, 6)$ form the vertices of a parallelogram. Let $P$ be the convex polygon whose vertices are the points in $S$. What is the area of $P$?

Let $ABCD$ be a parallelogram. Let $M$ be the midpoint of the arc $AC$ containing $B$ of the circumcircle of $ABC$ . Let $E$ be a point on segment $AD$ and $F$ a point on segment $CD$ such that $ME=MD=MF$. Show that $BMEF$ is cyclic. [i]Proposed by A. Tereshin[/i]

Given is a parallelogram $ABCD$ with $\angle A < 90^o$ and $|AB| < |BC|$. The angular bisector of angle $A$ intersects side $BC$ in $M$ and intersects the extension of $DC$ in $N$. Point $O$ is the centre of the circle through $M, C$, and $N$. Prove that $\angle OBC = \angle ODC$. [asy] unitsize (1.2 cm); pair A, B, C, D, M, N, O; A = (0,0); B = (2,0); D = (1,3); C = B + D - A; M = extension(A, incenter(A,B,D), B, C); N = extension(A, incenter(A,B,D), D, C); O = circumcenter(C,M,N); draw(D--A--B--C); draw(interp(D,N,-0.1)--interp(D,N,1.1)); draw(A--interp(A,N,1.1)); draw(circumcircle(M,C,N)); label("$\circ$", A + (0.45,0.15)); label("$\circ$", A + (0.25,0.35)); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$M$", M, SE); dot("$N$", N, dir(90)); dot("$O$", O, SE); [/asy]

$C_0:x^2+y^2=1,C_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}(a>b>0)$. Find all $(a,b)$ such that for any point $P$ on $C_1$, we can find a parallelogram with an apex $P$, and it is externally tangent to $C_0$, inscribed to $C_1$.

On sides $AB$ and $AC$ of an acute triangle $\Delta ABC$, with orthocenter $H$ and circumcenter $O$, are given points $P$ and $Q$ respectively such that $APHQ$ is a parallelogram. Prove the following equality: \begin{align*} \frac{PB\cdot PQ}{QA\cdot QO}=2 \end{align*}

Let $\Omega$ be the circumscribed circle of the acute triangle $ABC$ and $ D $ a point the small arc $BC$ of $\Omega$. Points $E$ and $ F $ are on the sides $ AB$ and $AC$, respectively, such that the quadrilateral $CDEF$ is a parallelogram. Point $G$ is on the small arc $AC$ such that lines $DC$ and $BG$ are parallel. Prove that the angles $GFC$ and $BAC$ are equal.

Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles. [asy] unitsize(1.5 cm); pair A, B, C, D, G, H; A = (0,0); B = (2,0); D = (0.5,1.5); C = B + D - A; G = reflect(A,B)*(C) + C - B; H = reflect(B,C)*(H) + A - B; draw(H--A--D--C--G); draw(interp(A,G,-0.1)--interp(A,G,1.1)); draw(interp(C,H,-0.1)--interp(C,H,1.1)); draw(D--G--H--cycle, dashed); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, E); dot("$D$", D, NW); dot("$G$", G, NE); dot("$H$", H, SE); [/asy]

On the sides of triangle $ABC$, isosceles right-angled triangles $AUB, CVB$, and $AWC$ are placed. These three triangles have their right angles at vertices $U, V$ , and $W$, respectively. Triangle $AUB$ lies completely inside triangle $ABC$ and triangles $CVB$ and $AWC$ lie completely outside $ABC$. See the figure. Prove that quadrilateral $UVCW$ is a parallelogram. [asy] import markers; unitsize(1.5 cm); pair A, B, C, U, V, W; A = (0,0); B = (2,0); C = (1.7,2.5); U = (B + rotate(90,A)*(B))/2; V = (B + rotate(90,C)*(B))/2; W = (C + rotate(90,A)*(C))/2; draw(A--B--C--cycle); draw(A--W, StickIntervalMarker(1,1,size=2mm)); draw(C--W, StickIntervalMarker(1,1,size=2mm)); draw(B--V, StickIntervalMarker(1,2,size=2mm)); draw(C--V, StickIntervalMarker(1,2,size=2mm)); draw(A--U, StickIntervalMarker(1,3,size=2mm)); draw(B--U, StickIntervalMarker(1,3,size=2mm)); draw(rightanglemark(A,U,B,5)); draw(rightanglemark(B,V,C,5)); draw(rightanglemark(A,W,C,5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, N); dot("$U$", U, NE); dot("$V$", V, NE); dot("$W$", W, NW); [/asy]

Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$

In an acute angled triangle $ABC$, $\angle A = 30^{\circ}$, $H$ is the orthocenter, and $M$ is the midpoint of $BC$. On the line $HM$, take a point $T$ such that $HM = MT$. Show that $AT = 2 BC$.

Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]

As shown in the figure, quadrilateral $ ABCD$ is inscribed in a circle with $ AC$ as its diameter, $ BD \perp AC,$ and $ E$ the intersection of $ AC$ and $ BD.$ Extend line segment $ DA$ and $ BA$ through $ A$ to $ F$ and $ G$ respectively, such that $ DG \parallel{} BF.$ Extend $ GF$ to $ H$ such that $ CH \perp GH.$ Prove that points $ B, E, F$ and $ H$ lie on one circle. [asy] defaultpen(linewidth(0.8)+fontsize(10));size(150); real a=4, b=6.5, c=9, d=a*c/b, g=14, f=sqrt(a^2+b^2)*sqrt(a^2+d^2)/g; pair E=origin, A=(0,a), B=(-b,0), C=(0,-c), D=(d,0), G=A+g*dir(B--A), F=A+f*dir(D--A), M=midpoint(G--C); path c1=circumcircle(A,B,C), c2=Circle(M, abs(M-G)); pair Hf=F+10*dir(G--F), H=intersectionpoint(F--Hf, c2); dot(A^^B^^C^^D^^E^^F^^G^^H); draw(c1^^c2^^G--D--C--A--G--F--D--B--A^^F--H--C--B--F); draw(H--B^^F--E^^G--C, linetype("2 2")); pair point= E; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G)); label("$H$", H, dir(point--H)); label("$E$", E, NE);[/asy]

A street has parallel curbs $ 40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $ 15$ feet and each stripe is $ 50$ feet long. Find the distance, in feet, between the stripes. $ \textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25$

Let be a convex quadrilateral $ ABCD $ and the points $ M,N,P,Q $ such that $ MAB\sim NBC\sim PCD\sim QDA. $ [b]a)[/b] Prove that $ ABCD $ is a parallelogram if and only if $ MNPQ $ is a parallelogram. [b]b)[/b] Show that if the diagonals of $ MNPQ $ are congruent and perpendicular, then the diagonals of $ ABCD $ are congruent and perpendicular, or $ MAB $ is a right isosceles triangle. [i]Nelu Chichirim[/i]

Let $ABCD$ be a cyclic quadrilateral and let $H_{A}, H_{B}, H_{C}, H_{D}$ be the orthocenters of the triangles $BCD$, $CDA$, $DAB$ and $ABC$ respectively. Show that the quadrilaterals $ABCD$ and $H_{A}H_{B}H_{C}H_{D}$ are congruent.

The point $P$, inside the triangle $[ABC]$, lies on the perpendicular bisector of $[AB]$. $Q$ and $R$ points, exterior to the triangle, they are such that $ [BPA], [BQC]$ and $[CRA]$ are similar triangles. Shows that $[PQCR]$ is a parallelogram. [img]https://cdn.artofproblemsolving.com/attachments/f/5/6e036b127f8a013794b8246cbb1544e7280d4a.png[/img]

Let $AC$ be the longer of the two diagonals of the parallelogram $ABCD$. Drop perpendiculars from $C$ to $AB$ and $AD$ extended. If $E$ and $F$ are the feet of these perpendiculars, prove that $$AB \cdot AE + AD \cdot AF = (AC)^2.$$

Given $ \triangle ABC$ with point $ D$ inside. Let $ A_0\equal{}AD\cap BC$, $ B_0\equal{}BD\cap AC$, $ C_0 \equal{}CD\cap AB$ and $ A_1$, $ B_1$, $ C_1$, $ A_2$, $ B_2$, $ C_2$ are midpoints of $ BC$, $ AC$, $ AB$, $ AD$, $ BD$, $ CD$ respectively. Two lines parallel to $ A_1A_2$ and $ C_1C_2$ and passes through point $ B_0$ intersects $ B_1B_2$ in points $ A_3$ and $ C_3$respectively. Prove that $ \frac{A_3B_1}{A_3B_2}\equal{}\frac{C_3B_1}{C_3B_2}$.