Let the incircle $k$ of the triangle $ABC$ touch its side $BC$ at $D$. Let the line $AD$ intersect $k$ at $L \neq D$ and denote the excentre of $ABC$ opposite to $A$ by $K$. Let $M$ and $N$ be the midpoints of $BC$ and $KM$ respectively. Prove that the points $B, C, N,$ and $L$ are concyclic.

Show that in a non-equilateral triangle, the following statements are equivalent: $(a)$ The angles of the triangle are in arithmetic progression. $(b)$ The common tangent to the Nine-point circle and the Incircle is parallel to the Euler Line.

Janabel has a device that, when given two distinct points $U$ and $V$ in the plane, draws the perpendicular bisector of $UV$. Show that if three lines forming a triangle are drawn, Janabel can mark the orthocenter of the triangle using this device, a pencil, and no other tools. [i]Proposed by Fedir Yudin.[/i]

Initially, there are three different points on the plane. Every minute, three points are chosen, for example $A$, $B$ and $C$, and a new point $D$ is generated which is symmetric to $A$ with respect to the perpendicular bisector of line segment $BC$. 24 hours later, it turns out that among all the points that were generated, there exist three collinear points. Prove that the three initial points were also collinear. (Author: V. Shmarov)

Starting from a given cyclic quadrilateral $\mathcal{Q}_0$, a sequence of quadrilaterals is constructed so that $\mathcal{Q}_{k + 1}$ is the circumscribed quadrilateral of $\mathcal{Q}_k$ for $k = 0,1,\dots$. The sequence terminates when a quadrilateral is reached that is not cyclic. (The circumscribed quadrilateral of a cylic quadrilateral $ABCD$ has sides that are tangent to the circumcircle of $ABCD$ at $A$, $B$, $C$ and $D$.) Prove that the sequence always terminates, except when $\mathcal{Q}_0$ is a square.

Assume that all angles of a triangle $ABC$ are acute. Let $D$ and $E$ be points on the sides $AC$ and $BC$ of the triangle such that $A, B, D,$ and $E$ lie on the same circle. Further suppose the circle through $D,E,$ and $C$ intersects the side $AB$ in two points $X$ and $Y$. Show that the midpoint of $XY$ is the foot of the altitude from $C$ to $AB$.

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]

In convex hexagon $ABCDEF$, $\angle A \cong \angle B$, $\angle C \cong \angle D$, and $\angle E \cong \angle F$. Prove that the perpendicular bisectors of $\overline{AB}$, $\overline{CD}$, and $\overline{EF}$ pass through a common point. [i]Proposed by Lewis Chen[/i]

Let $ABC$ be an acute triangle with $AC > AB > BC$. The perpendicular bisectors of $AC$ and $AB$ cut line $BC$ at $D$ and $E$ respectively. Let $P$ and $Q$ be points on lines $AC$ and $AB$ respectively, both different from $A$, such that $AB = BP$ and $AC = CQ$, and let $K$ be the intersection of lines $EP$ and $DQ$. Let $M$ be the midpoint of $BC$. Show that $\angle DKA = \angle EKM$.

Let $ABC$ be a scalene triangle with circumcenter $O$, and let $D$ and $E$ be points inside angle $\measuredangle BAC$ such that $A$ lies on line $DE$, and $\angle ADB=\angle CBA$ and $\angle AEC=\angle BCA$. Let $M$ be the midpoint of $BC$ and $K$ be a point such that $OK$ is perpendicular to $AO$ and $\angle BAK=\angle MAC$. Finally, let $P$ be the intersection of the perpendicular bisectors of $BD$ and $CE$. Show that $KO=KP$. [i]Proposed by Victor Domínguez[/i]

Let $ABCC_1B_1A_1$ be a convex hexagon such that $AB=BC$, and suppose that the line segments $AA_1, BB_1$, and $CC_1$ have the same perpendicular bisector. Let the diagonals $AC_1$ and $A_1C$ meet at $D$, and denote by $\omega$ the circle $ABC$. Let $\omega$ intersect the circle $A_1BC_1$ again at $E \neq B$. Prove that the lines $BB_1$ and $DE$ intersect on $\omega$.

Let $\triangle ABC$ be a scalene and acute triangle in which the angle at $A$ is second largest, $H$ is the orthocenter, and $k$ is the circumcircle with center $O$. Let the circumcircle of $\triangle AHO$ intersect the sides $AB$ and $AC$ again at $M$ and $N$, respectively, whereas the altitudes $CH$ and $BH$ intersect $k$ again at $K$ and $L$, respectively. Prove that the intersection of $KL$ and the perpendicular bisector of $AH$ is the orthocenter of $\triangle AMN$. Proposed by [i]Ilija Jovcevski[/i]

$BB_1$ is the angle bisector of $\triangle ABC$, and $I$ is its incenter. The perpendicular bisector of segment $AC$ intersects the circumcircle of $\triangle AIC$ at $D$ and $E$. Point $F$ is on the segment $B_1C$ such that $AB_1=CF$.Prove that the four points $B, D, E$ and $F$ are concyclic.

Suppose $\triangle ABC$ is scalene. $O$ is the circumcenter and $A'$ is a point on the extension of segment $AO$ such that $\angle BA'A = \angle CA'A$. Let point $A_1$ and $A_2$ be foot of perpendicular from $A'$ onto $AB$ and $AC$. $H_{A}$ is the foot of perpendicular from $A$ onto $BC$. Denote $R_{A}$ to be the radius of circumcircle of $\triangle H_{A}A_1A_2$. Similiarly we can define $R_{B}$ and $R_{C}$. Show that: \[\frac{1}{R_{A}} + \frac{1}{R_{B}} + \frac{1}{R_{C}} = \frac{2}{R}\] where R is the radius of circumcircle of $\triangle ABC$.

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 $PQ = r$.

Let $ABC$ be a triangle with $AB>AC$. Let $D$ be a point on side $AB$ such that $BD=AC$. Consider the circle $\gamma$ passing through point $D$ and tangent to side $AC$ at point $A$. Consider the circumscribed circle $\omega$ of the triangle $ABC$ that interesects the circle $\gamma$ at points $A$ and $E$. Prove that point $E$ is the intersection point of the perpendicular bisectors of line segments $BC$ and $AD$.

In triangle $ABC$, let $J$ be the center of the excircle tangent to side $BC$ at $A_{1}$ and to the extensions of the sides $AC$ and $AB$ at $B_{1}$ and $C_{1}$ respectively. Suppose that the lines $A_{1}B_{1}$ and $AB$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $DJ$. Determine the angles $\angle{BEA_{1}}$ and $\angle{AEB_{1}}$. [i]Proposed by Dimitris Kontogiannis, Greece[/i]

[b]i)[/b] Let $0<a<b$.Prove that amongst all triangles having base $a$ and perimeter $a+b$ the triangle having two sides(other than the base) equal to $\frac {b}{2}$ has the maximum area. [b]ii)[/b]Using $i)$ or otherwise, prove that amongst all quadrilateral having give perimeter the square has the maximum area.

Given a circle $(O)$ and two fixed points $B,C$ on $(O),$ and an arbitrary point $A$ on $(O)$ such that the triangle $ABC$ is acute. $M$ lies on ray $AB,$ $N$ lies on ray $AC$ such that $MA=MC$ and $NA=NB.$ Let $P$ be the intersection of $(AMN)$ and $(ABC),$ $P\ne A.$ $MN$ intersects $BC$ at $Q.$ a) Prove that $A,P,Q$ are collinear. b) $D$ is the midpoint of $BC.$ Let $K$ be the intersection of $(M,MA)$ and $(N,NA),$ $K\ne A.$ $d$ is the line passing through $A$ and perpendicular to $AK.$ $E$ is the intersection of $d$ and $BC.$ $(ADE)$ intersects $(O)$ at $F,$ $F\ne A.$ Prove that $AF$ passes through a fixed point.

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]

In a plane a circle with radius $R$ and center $w$ and a line $\Lambda$ are given. The distance between $w$ and $\Lambda$ is $d, d > R$. The points $M$ and $N$ are chosen on $\Lambda$ in such a way that the circle with diameter $MN$ is externally tangent to the given circle. Show that there exists a point $A$ in the plane such that all the segments $MN$ are seen in a constant angle from $A.$

Let $\triangle ABC$ be a triangle with circumcenter $O$. The perpendicular bisectors of the segments $OA,OB$ and $OC$ intersect the lines $BC,CA$ and $AB$ at $D,E$ and $F$, respectively. Prove that $D,E,F$ are collinear.

Given a triangle $ABC$, in which $\angle B> 90^o$. Perpendicular bisector of the side $AB$ intersects the side $AC$ at the point $M$, and the perpendicular bisector of the side $AC$ intersects the extension of the side $AB$ beyond the vertex $B$ at point $N$. It is known that the segments $MN$ and $BC$ are equal and intersect at right angles. Find the values ​​of all angles of triangle $ABC$.

In $\triangle ABC$, side $BC>AB$. Point $D$ lies on side $AC$ such that $\angle ABD=\angle CBD$. Points $Q,P$ lie on line $BD$ such that $AQ\bot BD$ and $CP\bot BD$. $M,E$ are the midpoints of side $AC$ and $BC$ respectively. Circle $O$ is the circumcircle of $\triangle PQM$ intersecting side $AC$ at $H$. Prove that $O,H,E,M$ lie on a circle.

Let AD be the altitude of a triangle ABC and E , F be the incenters of the triangle ABD and ACD , respectively. line EF meets AB and AC at K and L. prove tht AK=AL if and only if AB=AC or A=90