Given is a triangle $ABC$ with barycenter $G$ and circumcenter $O$.The perpendicular bisectors of $GA,GB,GC$ intersect at $A_1,B_1,C_1$.Show that $O$ is the barycenter of $\triangle{A_1B_1C_1}$.

Let $ABC$ be a triangle, and $M$ and $N$ variable points on $AB$ and $AC$ respectively, such that both $M$ and $N$ do not lie on the vertices, and also, $AM \times MB = AN \times NC$. Prove that the perpendicular bisector of $MN$ passes through a fixed point.

The inscribed circle $\omega$ of the non-isosceles acute-angled triangle $ABC$ touches the side $BC$ at the point $D$. Suppose that $I$ and $O$ are the centres of inscribed circle and circumcircle of triangle $ABC$ respectively. The circumcircle of triangle $ADI$ intersects $AO$ at the points $A$ and $E$. Prove that $AE$ is equal to the radius $r$ of $\omega$.

$AB$ is a diameter of a circle with center $O$. Let $C$ and $D$ be two different points on the circle on the same side of $AB$, and the lines tangent to the circle at points $C$ and $D$ meet at $E$. Segments $AD$ and $BC$ meet at $F$. Lines $EF$ and $AB$ meet at $M$. Prove that $E,C,M$ and $D$ are concyclic.

A point $ P$ is randomly selected from the rectangular region with vertices $ (0, 0)$, $ (2, 0)$, $ (2, 1)$, $ (0, 1)$. What is the probability that $ P$ is closer to the origin than it is to the point $ (3, 1)$? $ \textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{4}{5} \qquad \textbf{(E)}\ 1$

Let $A$ and $B$ are two points on a plane, and let $M$ be the midpoint of $AB$. Let $r$ be a line and let $R$ and $S$ be the projections of $A$ and $B$ onto $r$. Assuming that $A$, $M$, and $R$ are not collinear, prove that the circumcircle of triangle $AMR$ has the same radius as the circumcircle of $BSM$.

Triangle $ABC$ is such that $AB < AC$. The perpendicular bisector of side $BC$ intersects lines $AB$ and $AC$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $ABC$, and let $M$ and $N$ be the midpoints of segments $BC$ and $PQ$, respectively. Prove that lines $HM$ and $AN$ meet on the circumcircle of $ABC$.

Let there be an acute triangle $ABC$ with orthocenter $H$. Let $BH, CH$ hit the circumcircle of $\triangle ABC$ at $D, E$. Let $P$ be a point on $AB$, between $B$ and the foot of the perpendicular from $C$ to $AB$. Let $PH \cap AC = Q$. Now $\triangle AEP$'s circumcircle hits $CH$ at $S$, $\triangle ADQ$'s circumcircle hits $BH$ at $R$, and $\triangle AEP$'s circumcircle hits $\triangle ADQ$'s circumcircle at $J (\not=A)$. Prove that $RS$ is the perpendicular bisector of $HJ$.

Given is an acute triangle $ABC$ $\left(AB<AC<BC\right)$,inscribed in circle $c(O,R)$.The perpendicular bisector of the angle bisector $AD$ $\left(D\in BC\right)$ intersects $c$ at $K,L$ ($K$ lies on the small arc $\overarc{AB}$).The circle $c_1(K,KA)$ intersects $c$ at $T$ and the circle $c_2(L,LA)$ intersects $c$ at $S$.Prove that $\angle{BAT}=\angle{CAS}$. [hide=Diagram][asy]import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.94236331697463, xmax = 15.849400903703716, ymin = -5.002235438802758, ymax = 7.893104843949444; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen qqqqtt = rgb(0.,0.,0.2); draw((1.8318261909633622,3.572783369254345)--(0.,0.)--(6.,0.)--cycle, aqaqaq); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-117.14497824050169,-101.88970202103212)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-55.85706977865775,-40.60179355918817)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); /* draw figures */ draw((1.8318261909633622,3.572783369254345)--(0.,0.), uququq); draw((0.,0.)--(6.,0.), uququq); draw((6.,0.)--(1.8318261909633622,3.572783369254345), uququq); draw(circle((3.,0.7178452373968209), 3.0846882800136055)); draw((2.5345020274407277,0.)--(1.8318261909633622,3.572783369254345)); draw(circle((-0.01850947366601585,1.3533783539547308), 2.889550258039566)); draw(circle((5.553011501106743,2.4491551634556963), 3.887127532933951)); draw((-0.01850947366601585,1.3533783539547308)--(5.553011501106743,2.4491551634556963), linetype("2 2")); draw((1.8318261909633622,3.572783369254345)--(0.7798408954511686,-1.423695174396108)); draw((1.8318261909633622,3.572783369254345)--(5.22015910454883,-1.4236951743961088)); /* dots and labels */ dot((1.8318261909633622,3.572783369254345),linewidth(3.pt) + dotstyle); label("$A$", (1.5831274347452782,3.951671933606579), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.6,0.05), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.188606107156787,0.07450151636712989), NE * labelscalefactor); dot((2.5345020274407277,0.),linewidth(3.pt) + dotstyle); label("$D$", (2.3,-0.7), NE * labelscalefactor); dot((-0.01850947366601585,1.3533783539547308),linewidth(3.pt) + dotstyle); label("$K$", (-0.3447473583572136,1.6382221818835927), NE * labelscalefactor); dot((5.553011501106743,2.4491551634556963),linewidth(3.pt) + dotstyle); label("$L$", (5.631664500260511,2.580738747400365), NE * labelscalefactor); dot((0.7798408954511686,-1.423695174396108),linewidth(3.pt) + dotstyle); label("$T$", (0.5977692071595602,-1.960477431907719), NE * labelscalefactor); dot((5.22015910454883,-1.4236951743961088),linewidth(3.pt) + dotstyle); label("$S$", (5.160406217502124,-1.8747941077698307), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

Let the circles $\omega$ and $\omega'$ intersect in points $A$ and $B$. The tangent to circle $\omega$ at $A$ intersects $\omega'$ at $C$ and the tangent to circle $\omega'$ at $A$ intersects $\omega$ at $D$. Suppose that the internal bisector of $\angle CAD$ intersects $\omega$ and $\omega'$ at $E$ and $F$, respectively, and the external bisector of $\angle CAD$ intersects $\omega$ and $\omega'$ at $X$ and $Y$, respectively. Prove that the perpendicular bisector of $XY$ is tangent to the circumcircle of triangle $BEF$. [i]Proposed by Mahdi Etesami Fard[/i]

Six different points $A, B, C, D, E, F$ are marked on the plane, no four of them lie on one circle and no two segments with ends at these points lie on parallel lines. Let $P, Q,R$ be the points of intersection of the perpendicular bisectors to pairs of segments $(AD, BE)$, $(BE, CF)$ ,$(CF, DA)$ respectively, and $P', Q' ,R'$ are points the intersection of the perpendicular bisectors to the pairs of segments $(AE, BD)$, $(BF, CE)$ , $(CA, DF)$ respectively. Show that $P \ne P', Q \ne Q', R \ne R'$, and prove that the lines $PP', QQ'$ and $RR'$ intersect at one point or are parallel.

A convex quadrilateral $ABCD$ has an incircle. In each corner a circle is inscribed that also externally touches the two circles inscribed in the adjacent corners. Show that at least two circles have the same size.

Let $ABC$ be an acute-angled triangle. The line through $C$ perpendicular to $AC$ meets the external angle bisector of $\angle ABC$ at $D$. Let $H$ be the foot of the perpendicular from $D$ onto $BC$. The point $K$ is chosen on $AB$ so that $KH \parallel AC$. Let $M$ be the midpoint of $AK$. Prove that $MC = MB + BH$. Giorgi Arabidze, Georgia,

Let $ABCD$ be a parallelogram with $AB>BC$ and $\angle DAB$ less than $\angle ABC$. The perpendicular bisectors of sides $AB$ and $BC$ intersect at the point $M$ lying on the extension of $AD$. If $\angle MCD=15^{\circ}$, find the measure of $\angle ABC$

Let $ABC$ be a triangle with $\angle C = 90^\circ$ and $CA \neq CB$. Let $CH$ be an altitude and $CL$ be an interior angle bisector. Show that for $X \neq C$ on the line $CL$, we have $\angle XAC \neq \angle XBC$. Also show that for $Y \neq C$ on the line $CH$ we have $\angle YAC \neq \angle YBC$. [i]Bulgaria[/i]

$A,B,C$ are points that three complex numbers $z_0=a\text{i},z_1=\frac{1}{2}+b\text{i},z_2=1+c\text{i}(a,b,c\in\mathbb{R})$ refer to on complex plane (not collinear). Prove that curve $Z=Z_0\cos^4t+2Z_1\cos^2t\sin^2t+Z_2\sin^4t(t\in\mathbb{R})$ has only one common point with the perpendicular bisector of $AC$, and find the point.

Let $\Gamma$ be a circle with radius $r$. Let $A$ and $B$ be distinct points on $\Gamma$ such that $AB < \sqrt{3}r$. Let the circle with centre $B$ and radius $AB$ meet $\Gamma$ again at $C$. Let $P$ be the point inside $\Gamma$ such that triangle $ABP$ is equilateral. Finally, let the line $CP$ meet $\Gamma$ again at $Q$. Prove that $PQ = r$.

The incircle of triangle $ABC$ is tangent to $BC,CA,AB$ at $D,E,F$ respectively. Consider the triangle formed by the line joining the midpoints of $AE,AF$, the line joining the midpoints of $BF,BD$, and the line joining the midpoints of $CD,CE$. Prove that the circumcenter of this triangle coincides with the circumcenter of triangle $ABC$.

Consider a cyclic quadrilateral $ABCD$, and let $S$ be the intersection of $AC$ and $BD$. Let $E$ and $F$ the orthogonal projections of $S$ on $AB$ and $CD$ respectively. Prove that the perpendicular bisector of segment $EF$ meets the segments $AD$ and $BC$ at their midpoints.

Let $ABCD$ be a convex quadrilateral with $\angle CBD = 2 \angle ADB$, $\angle ABD = 2 \angle CDB$ and $AB = CB$. Prove that $AD = CD$.

Triangle $ABC$ is such that $AB < AC$. The perpendicular bisector of side $BC$ intersects lines $AB$ and $AC$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $ABC$, and let $M$ and $N$ be the midpoints of segments $BC$ and $PQ$, respectively. Prove that lines $HM$ and $AN$ meet on the circumcircle of $ABC$.

Diagonals $AC$ and $BD$ of a trapezoid $ABCD$ meet at $P$. The circumcircles of triangles $ABP$ and $CDP$ intersect the line $AD$ for the second time at points $X$ and $Y$ respectively. Let $M$ be the midpoint of segment $XY$. Prove that $BM = CM$.

In the figure of [url]http://www.artofproblemsolving.com/Forum/download/file.php?id=50643&mode=view[/url] $\odot O_1$ and $\odot O_2$ intersect at two points $A$, $B$. The extension of $O_1A$ meets $\odot O_2$ at $C$, and the extension of $O_2A$ meets $\odot O_1$ at $D$, and through $B$ draw $BE \parallel O_2A$ intersecting $\odot O_1$ again at $E$. If $DE \parallel O_1A$, prove that $DC \perp CO_2$.

As shown in the following figure, point $ P$ lies on the circumcicle of triangle $ ABC.$ Lines $ AB$ and $ CP$ meet at $ E,$ and lines $ AC$ and $ BP$ meet at $ F.$ The perpendicular bisector of line segment $ AB$ meets line segment $ AC$ at $ K,$ and the perpendicular bisector of line segment $ AC$ meets line segment $ AB$ at $ J.$ Prove that \[ \left(\frac{CE}{BF} \right)^2 \equal{} \frac{AJ \cdot JE}{AK \cdot KF}.\]

Let $ABC$ be a triangle satisfying $AB<AC$. Let $M$ be the midpoint of $BC$. A point $P$ lies on the segment $AB$ with $AP>PB$. A point $Q$ lies on the segment $AC$ with $\angle MPA = \angle AQM$. The perpendicular bisectors of $BC$ and $PQ$ intersect at $S$. Prove that $\angle BAC + \angle QSP = \angle QMP$.