Let $\triangle ABC$ have $M$ as the midpoint of $BC$ and let $P$ and $Q$ be the feet of the altitudes from $M$ to $AB$ and $AC$ respectively. Find $\angle BAC$ if $[MPQ]=\frac{1}{4}[ABC]$ and $P$ and $Q$ lie on the segments $AB$ and $AC$.

Let $ABC$ be an acute triangle with $AC> AB$. be $\Gamma$ the circumcircle circumscribed to the triangle $ABC$ and $D$ the midpoint of the smallest arc $BC$ of this circle. Let $E$ and $F$ points of the segments $AB$ and $AC$ respectively such that $AE = AF$. Let $P \neq A$ be the second intersection point of the circumcircle circumscribed to $AEF$ with $\Gamma$. Let $G$ and $H$ be the intersections of lines $PE$ and $PF$ with $\Gamma$ other than $P$, respectively. Let $J$ and $K$ be the intersection points of lines $DG$ and $DH$ with lines $AB$ and $AC$ respectively. Show that the $JK$ line passes through the midpoint of $BC$

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

Let $A B C$ be an acute-angled triangle of circumcenter $O$ and orthocenter $H$. Let $M$ be the midpoint of $BC, N$ be the symmetric of $H$ with respect to $A, P$ be the midpoint of $NM$ and $X$ be a point on the line A H such that $MX$ is parallel to $CH$. Prove that $BX$ and $OP$ are perpendicular.

Let $ABCD$ be a cyclic quadrilateral, and let diagonals $AC$ and $BD$ intersect at $X$.Let $C_1,D_1$ and $M$ be the midpoints of segments $CX,DX$ and $CD$, respecctively. Lines $AD_1$ and $BC_1$ intersect at $Y$, and line $MY$ intersects diagonals $AC$ and $BD$ at different points $E$ and $F$, respectively. Prove that line $XY$ is tangent to the circle through $E,F$ and $X$.

Quadrilateral $ABCD$ is circumscribed around a circle with center $I$. Points $M$ and $N$ are the midpoints of diagonals $AC$ and $BD$. Prove that $ABCD$ is cyclic quadrilateral if and only if $IM : AC = IN : BD$. [i]Nikolai Beluhov and Aleksey Zaslavsky[/i]

Let $C$ be an interior point of line segment $AB$. Equilateral triangles $ADC$ and $CEB$ are constructed to the same side from $AB$. Find all points which can be the midpoint of the segment $DE$.

Let $ABC$ be an acute triangle with circumcenter $O$. Points $E$ and $F$ are chosen on segments $OB$ and $OC$ such that $BE = OF$. If $M$ is the midpoint of the arc $EOA$ and $N$ is the midpoint of the arc $AOF$, prove that $\sphericalangle ENO + \sphericalangle OMF = 2 \sphericalangle BAC$.

Let $D$ and $E$ be the midpoints of sides $[AB]$ and $[AC]$ of the triangle $ABC$. The circle of diameter $[AB]$ intersects the line $DE$ on the opposite side of $AB$ than $C$, in $X$. The circle of diameter $[AC]$ intersects $DE$ on the opposite side of $AC$ than $B$ in $Y$ . Let $T$ be the intersection of $BX$ and $CY$. Prove that the orthocenter of triangle $XY T$ lies on $BC$.

Let $O$ be the center of the circle $\omega$ circumscribed around the acute-angled triangle $\vartriangle ABC$, and $W$ be the midpoint of the arc $BC$ of the circle $\omega$, which does not contain the point $A$, and $H$ be the point of intersection of the heights of the triangle $\vartriangle ABC$. Find the angle $\angle BAC$, if $WO = WH$. (O. Clurman)

A square $ABCD$ is given. Two circles are inscribed into angles $A$ and $B$, and the sum of their diameters is equal to the sidelength of the square. Prove that one of their common tangents passes through the midpoint of $AB$.

Juku invented an apparatus that can divide any segment into three equal segments. How can you find the midpoint of any segment, using only the Juku made, a ruler and pencil?

Let $ABC$ be an isosceles triangle ($AB = AC$), $D$ be the midpoint of $BC$, and $M$ be the midpoint of $AD$. On the segment $BM$ take a point $N$ such that $\angle BND = 90^o$. Find the angle $ANC$.

The points $E$ and $F$ divide the diagonal $BD$ of the convex quadrilateral $ABCD$ into three equal parts, i.e. $| BE | = | EF | = | F D |$. Line $AE$ interects side $BC$ at $X$ and line $AF$ intersects $DC$ at $Y$. Prove that: a) if $ABCD$ is parallelogram then $X ,Y$ are the midpoints of $BC, DC$, respectively, b) if the points $X , Y$ are the midpoints of $BC, DC$, respectively , then $ABCD$ is parallelogram

In making Euclidean constructions in geometry it is permitted to use a ruler and a pair of compasses. In the constructions considered in this question no compasses are permitted, but the ruler is assumed to have two parallel edges, which can be used for constructing two parallel lines through two given points whose distance is at least equal to the breadth of the rule. Then the distance between the parallel lines is equal to the breadth of the ruler. Carry through the following constructions with such a ruler. Construct: [b]a)[/b] The bisector of a given angle. [b]b)[/b] The midpoint of a given rectilinear line segment. [b]c)[/b] The center of a circle through three given non-collinear points. [b]d)[/b] A line through a given point parallel to a given line.

A convex hexagon is circumscribed about a circle of radius $1$. Consider the three segments joining the midpoints of its opposite sides. Find the greatest real number $r$ such that the length of at least one segment is at least $r.$

Let $AA_1$ and $BB_1$ be the altitudes of acute angled triangle $ABC$. Points $K$ and $M$ are midpoints of $AB$ and $A_1B_1$ respectively. Segments $AA_1$ and $KM$ intersect at point $L$. Prove that points $A$, $K$, $L$ and $B_1$ are noncyclic. [I]Proposed by S. Berlov[/i]

Let $ABC$ be an acute triangle and $M$ be the midpoint of $AB$. A circle through the points $B$ and $C$ intersects the segments $CM$ and $BM$ at points $P$ and $Q$ respectively. Point $K$ is symmetric to $P$ with respect to point $M$. The circumcircles of $\triangle AKM$ and $\triangle CQM$ intersect for the second time at $X$. The circumcircles of $\triangle AMC$ and $\triangle KMQ$ intersect for the second time at $Y$. The segments $BP$ and $CQ$ intersect at point $T$. Prove that the line $MT$ is tangent to the circumcircle of $\triangle MXY$.

The circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$, point $M$ is the midpoint of $AB$. On line $AB$ select points $S_1$ and $S_2$. Let $S_1X_1$ and $S_1Y_1$ be tangents drawn from $S_1$ to circle $\omega_1$, similarly $S_2X_2$ and $S_2Y_2$ are tangents drawn from $S_2$ to circles $\omega_2$. Prove that if the point $M$ lies on the line $X_1X_2$, then it also lies on the line $Y_1Y_2$.

Let \(ABC\) be an isosceles triangle with \(AB = BC\). Let \(D\) be a point on segment \(AB\), \(E\) be a point on segment \(BC\), and \(P\) be a point on segment \(DE\) such that \(AD = DP\) and \(CE = PE\). Let \(M\) be the midpoint of \(DE\). The line parallel to \(AB\) through \(M\) intersects \(AC\) at \(X\) and the line parallel to \(BC\) through \(M\) intersects \(AC\) at \(Y\). The lines \(DX\) and \(EY\) intersect at \(F\). Prove that \(FP\) is perpendicular to \(DE\).

Let $a, b$ be two lines intersecting each other at $O$. Point $M$ is not on either $a$ or $b$. A variable circle $(C)$ passes through $O,M$ intersecting $a, b$ at $A,B$ respectively, distinct from $O$. Prove that the midpoint of $AB$ is on a fixed line. Hạ Vũ Anh

Consider the equilateral triangle $ABC$ with center of gravity $G$. Let $M$ be a point, inside the triangle and $O$ be the midpoint of the segment $MG$. Three segments go through $M$, each parallel to one side of the triangle and with the ends on the other two sides of the given triangle. a) Show that $O$ is at equal distances from the midpoints of the three segments considered. b) Show that the midpoints of the three segments are the vertices of an equilateral triangle.

Let $XOY$ be a triangle with $\angle XOY = 90^o$. Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$, respectively. Suppose that $XN = 19$ and $YM =22$. What is $XY$?

Let $ABC$ be a triangle and $D, E, F$ the midpoints of the sides $BC, CA, AB$ respectively. Show that $\angle DAC = \angle ABE$ if and only if $\angle AFC = \angle BDA$

Let $\Delta ABC$ be a triangle with orthocenter $H$ and midpoints $M_a,M_b$, and $M_c$ of $BC$, $AC$, and $AB$ respectively. A circle with center $H$ intersects the lines $M_bM_a$, $M_bM_c$, and $M_cM_a$ in points $U_1,U_2,V_1,V_2,W_1,W_2$ respectively. Prove that $CU_1=CU_2=AV_1=AV_2=BW_1=BW_2$.