Let $H$ be the orthocenter of a triangle $ABC$, and let $D$ be the midpoint of the segment $AH$. The altitude $BH$ of triangle $ABC$ intersects the perpendicular to the line $AB$ through the point $A$ at the point $M$. The altitude $CH$ of triangle $ABC$ intersects the perpendicular to the line $CA$ through the point $A$ at the point $N$. The perpendicular bisector of the segment $AB$ intersects the perpendicular to the line $BC$ through the point $B$ at the point $U$. The perpendicular bisector of the segment $CA$ intersects the perpendicular to the line $BC$ through the point $C$ at the point $V$. Finally, let $E$ be the midpoint of the side $BC$ of triangle $ABC$. Prove that the points $D$, $M$, $N$, $U$, $V$ all lie on one and the same perpendicular to the line $AE$. [i]Extensions.[/i] In other words, we have to show that the points $M$, $N$, $U$, $V$ lie on the perpendicular to the line $AE$ through the point $D$. Additionally, one can find two more points on this perpendicular: [b](a)[/b] The nine-point circle of triangle $ABC$ is known to pass through the midpoint $E$ of its side $BC$. Let $D^{\prime}$ be the point where this nine-point circle intersects the line $AE$ apart from $E$. Then, the point $D^{\prime}$ lies on the perpendicular to the line $AE$ through the point $D$. [b](b)[/b] Let the tangent to the circumcircle of triangle $ABC$ at the point $A$ intersect the line $BC$ at a point $X$. Then, the point $X$ lies on the perpendicular to the line $AE$ through the point $D$. [i]Comment.[/i] The actual problem was created by Victor Thébault around 1950 (cf. Hyacinthos messages #1102 and #1551). The extension [b](a)[/b] initially was a (pretty trivial) lemma in Thébault's solution of the problem. Extension [b](b)[/b] is rather new; in the form "prove that $X\in UV$", it was [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3659]proposed by Valentin Vornicu for the Balkan MO 2003[/url], however it circulated in the Hyacinthos newsgroup before (Hyacinthos messages #7240 and #7242), where different solutions of the problem were discussed as well. Hereby, "Hyacinthos" always refers to the triangle geometry newsgroup "Hyacinthos", which can be found at http://groups.yahoo.com/group/Hyacinthos . I proposed the problem for the QEDMO math fight wishing to draw some attention to it. It has a rather short and elementary solution, by the way (without using radical axes or inversion like the standard solutions). Darij

Consider an arbitrary equilateral triangle $KLM$, whose vertices $K, L$ and $M$ lie on the sides $AB, BC$ and $CD$, respectively, of a given square $ABCD$. Find the locus of the midpoints of the sides $KL$ of all such triangles $KLM$.

In a triangle $ABC$, the perpendicular bisectors of sides $AB$ and $AC$ intersect $BC$ at $X$ and $Y$. Prove that $BC = XY$ if and only if $\tan B\tan C = 3$ or $\tan B\tan C = -1$.

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 $\angle ABC$

Circles $\Gamma_1$ and $\Gamma_2$ have centers $O_1$ and $O_2$ and intersect at $P$ and $Q$. A line through $P$ intersects $\Gamma_1$ and $\Gamma_2$ at $A$ and $B$, respectively, such that $AB$ is not perpendicular to $PQ$. Let $X$ be the point on $PQ$ such that $XA=XB$ and let $Y$ be the point within $AO_1 O_2 B$ such that $AYO_1$ and $BYO_2$ are similar. Prove that $2\angle{O_1 AY}=\angle{AXB}$. [i]Author: Matthew Brennan[/i]

Let $ABC$ be a triangle such that $ABC$ isn't a isosceles triangle. $(I)$ is incircle of triangle touches $BC,CA,AB$ at $D,E,F$ respectively. The line through $E$ perpendicular to $BI$ cuts $(I)$ again at $K$. The line through $F$ perpendicular to $CI$ cuts $(I)$ again at $L$.$J$ is midpoint of $KL$. [b]a)[/b] Prove that $D,I,J$ collinear. [b]b)[/b] $B,C$ are fixed points,$A$ is moved point such that $\frac{AB}{AC}=k$ with $k$ is constant.$IE,IF$ cut $(I)$ again at $M,N$ respectively.$MN$ cuts $IB,IC$ at $P,Q$ respectively. Prove that bisector perpendicular of $PQ$ through a fixed point.

$AB$ 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.

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 $AB$.

As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$. [asy]import graph; size(5.55cm); pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.76,xmax=4.8,ymin=-3.69,ymax=3.71; pen zzttqq=rgb(0.6,0.2,0), wwwwqq=rgb(0.4,0.4,0), qqwuqq=rgb(0,0.39,0); pair A=(-2,2.5), B=(-3,-1.5), C=(2,-1.5), I=(-1.27,-0.15), D=(-2.58,0.18), O=(-0.5,-2.92); D(A--B--C--cycle,zzttqq); D(arc(D,0.25,-104.04,-56.12)--(-2.58,0.18)--cycle,qqwuqq); D(arc((-0.31,0.81),0.25,-92.92,-45)--(-0.31,0.81)--cycle,qqwuqq); D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR(I,1.35),linewidth(1.2)+dotted+wwwwqq); D(CR(O,2.87),linetype("2 2")+blue); D(D--O); D((-0.31,0.81)--O); D(A); D(B); D(C); D(I); D(D); D((-0.31,0.81)); D(O); MP( "A", A, dir(110)); MP("B", B, dir(140)); D("C", C, dir(20)); D("D", D, dir(150)); D("E", (-0.31, 0.81), dir(60)); D("O", O, dir(290)); D("I", I, dir(100)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

Let $ P$ be the the isogonal conjugate of $ Q$ with respect to triangle $ ABC$, and $ P,Q$ are in the interior of triangle $ ABC$. Denote by $ O_{1},O_{2},O_{3}$ the circumcenters of triangle $ PBC,PCA,PAB$, $ O'_{1},O'_{2},O'_{3}$ the circumcenters of triangle $ QBC,QCA,QAB$, $ O$ the circumcenter of triangle $ O_{1}O_{2}O_{3}$, $ O'$ the circumcenter of triangle $ O'_{1}O'_{2}O'_{3}$. Prove that $ OO'$ is parallel to $ PQ$.

Let $ABC$ be acute-angled triangle . A circle $\omega_1(O_1,R_1)$ passes through points $B$ and $C$ and meets the sides $AB$ and $AC$ at points $D$ and $E$ ,respectively . Let $\omega_2(O_2,R_2)$ be the circumcircle of triangle $ADE$ . Prove that $O_1O_2$ is equal to the circumradius of triangle $ABC$ .

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.

A circle with center $O$ and a point $A$ in this circle are given. Let $P_{B}$ is the intersection point of $[AB]$ and the internal bisector of $\angle AOB$ where $B$ is a point on the circle such that $B$ doesn't lie on the line $OA$, Find the locus of $P_{B}$ as $B$ varies.

Let $ D$ be an interior point of the side $ BC$ of a triangle $ ABC$, and let $ O_1$ and $ O_2$ be the circumcenters of triangles $ ABD$ and $ ADC$. The perpendicular bisector of the side $ AC$ meets the line $ AO_1$ at $ E$, and the perpendicular bisector of the side $ AB$ meets the line $ AO_2$ at $ F$. Prove that the bisectors of the angles $ \angle O_1EO_2$ and $ \angle O_1FO_2$ are orthogonal.

Let $ A, B$ and $ C$ be non-collinear points. Prove that there is a unique point $ X$ in the plane of $ ABC$ such that \[ XA^2 \plus{} XB^2 \plus{} AB^2 \equal{} XB^2 \plus{} XC^2 \plus{} BC^2 \equal{} XC^2 \plus{} XA^2 \plus{} CA^2.\]