Let $ABCD$ be a parallelogram, such that the point $M$ is the midpoint of the side $CD$ and lies on the bisector of the angle $\angle BAD$. Prove that $\angle AMB = 90^o$.

Given a triangle $ABC$. Let $S$ be the circle passing through $C$, centered at $A$. Let $X$ be a variable point on $S$ and let $K$ be the midpoint of the segment $CX$ . Find the locus of the midpoints of $BK$, when $X$ moves along $S$. (I. Gorodnin)

In a triangle $ABC$ two points, $C_1$ and $A_1$ are marked on the sides $AB$ and $BC$ respectively (the points do not coincide with the vertices). Let $K$ be the midpoint of $A_1C_1$ and $I$ be the incentre of the triangle $ABC$. Given that the quadrilateral $A_1BC_1I$ is cyclic, prove that the angle $AKC$ is obtuse.

The sides $AB, BC, CD$ and $DA$ of the convex quadrilateral $ABCD$ have midpoints $E, F, G$ and $H$. Prove that the triangles $EFB, FGC, GHD$ and $HEA$ can be put together into a parallelogram equal to $EFGH$.

In triangle $ABC$, point $M$ is the midpoint of $BC, P$ is the intersection point of the tangents at points $B$ and $C$ of the circumscribed circle, $N$ is the midpoint of the segment $MP$. The segment $AN$ intersects the circumscribed circle at point $Q$. Prove that $\angle PMQ = \angle MAQ$.

Let $ABC$ be a triangle for which $AB \neq AC$. Points $K$, $L$, $M$ are the midpoints of the sides $BC$, $CA$, $AB$. The incircle of $ABC$ with center $I$ is tangent to $BC$ in $D$. A line passing through the midpoint of $ID$ perpendicular to $IK$ meets the line $LM$ in $P$. Prove that $\angle PIA = 90 ^{\circ}$.

We say that $M$ is the midpoint of the open polygonal $XYZ$, formed by the segments $XY, YZ$, if $M$ belongs to the polygonal and divides its length by half. Let $ABC$ be a acute triangle with orthocenter $H$. Let $A_1, B_1,C_1,A_2, B_2,C_2$ be the midpoints of the open polygonal $CAB, ABC, BCA, BHC, CHA, AHB$, respectively. Show that the lines $A_1 A_2, B_1B_2$ and $C_1C_2$ are concurrent.

Fix a triangle $ABC$. Let $\Gamma_1$ the circle through $B$, tangent to edge in $A$. Let $\Gamma_2$ the circle through C tangent to edge $AB$ in $A$. The second intersection of $\Gamma_1$ and $\Gamma_2$ is denoted by $D$. The line $AD$ has second intersection $E$ with the circumcircle of $\vartriangle ABC$. Show that $D$ is the midpoint of the segment $AE$.

Fix a triangle $ABC$. Let $\Gamma_1$ the circle through $B$, tangent to edge in $A$. Let $\Gamma_2$ the circle through C tangent to edge $AB$ in $A$. The second intersection of $\Gamma_1$ and $\Gamma_2$ is denoted by $D$. The line $AD$ has second intersection $E$ with the circumcircle of $\vartriangle ABC$. Show that $D$ is the midpoint of the segment $AE$.

Three chords $AB, CD$ and $EF$ of a circle intersect at the midpoint $M$ of $AB$. Show that if $CE$ produced and $DF$ produced meet the line $AB$ at the points $P$ and $Q$ respectively, then $M$ is also the midpoint of $PQ$.

Let $ABCD$ be a convex quadrilateral, and $P$ be a point in its interior. The projections of $P$ on the sides of the quadrilateral lie on a circle with center $O$. Show that $O$ lies on the line through the midpoints of $AC$ and $BD$.

In a triangle $ABC$ let $A_1$ be the midpoint of side $BC$. Draw circles with centers $A, A1$ and radii $AA_1, BC$ respectively and let $A'A''$ be their common chord. Similarly denote the segments $B'B''$ and $C'C''$. Show that lines $A'A'', B'B'''$ and $C'C''$ are concurrent.

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

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?

Given two distinct points $A_1,A_2$ in the plane, determine all possible positions of a point $A_3$ with the following property: There exists an array of (not necessarily distinct) points $P_1,P_2,...,P_n$ for some $n \ge 3$ such that the segments $P_1P_2,P_2P_3,...,P_nP_1$ have equal lengths and their midpoints are $A_1, A_2, A_3, A_1, A_2, A_3, ...$ in this order.

Let $P$ be an interior point of a circle $\Gamma$ centered at $O$ where $P \ne O$. Let $A$ and $B$ be distinct points on $\Gamma$. Lines $AP$ and $BP$ meet $\Gamma$ again at $C$ and $D$, respectively. Let $S$ be any interior point on line segment $PC$. The circumcircle of $\vartriangle ABS$ intersects line segment $PD$ at $T$. The line through $S$ perpendicular to $AC$ intersects $\Gamma$ at $U$ and $V$ . The line through $T$ perpendicular to $BD$ intersects $\Gamma$ at $X$ and $Y$ . Let $M$ and $N$ be the midpoints of $UV$ and $XY$ , respectively. Let $AM$ and $BN$ meet at $Q$. Suppose that $AB$ is not parallel to $CD$. Show that $P, Q$, and $O$ are collinear if and only if $S$ is the midpoint of $PC$.

Equilateral triangles $ABC_1, BCA_1, CAB_1$ are built on the sides of the triangle $ABC$ to the outside. On the segment $A_1B_1$ to the outer side of the triangle $A_1B_1C_1$, an equilateral triangle $A_1B_1C_2$ is constructed. Prove that $C$ is the midpoint of the segment $C_1C_2$. (A. Zaslavsky)

Let $ABC$ be a triangle. Its incircle meets the sides $BC, CA$ and $AB$ in the points $D, E$ and $F$, respectively. Let $P$ denote the intersection point of $ED$ and the line perpendicular to $EF$ and passing through $F$, and similarly let $Q$ denote the intersection point of $EF$ and the line perpendicular to $ED$ and passing through $D$. Prove that $B$ is the mid-point of the segment $PQ$. Proposed by Karl Czakler

Let $ABCD$ be a trapezoid with $AB // CD$, $2|AB| = |CD|$ and $BD \perp BC$. Let $M$ be the midpoint of $CD$ and let $E$ be the intersection $BC$ and $AD$. Let $O$ be the intersection of $AM$ and $BD$. Let $N$ be the intersection of $OE$ and $AB$. (a) Prove that $ABMD$ is a rhombus. (b) Prove that the line $DN$ passes through the midpoint of the line segment $BE$.

The sides $BC, CA$, and $AB$ of a triangle $ABC$ are tangent to a circle at points $X, Y, Z$ respectively. Show that the center of such a circle is on the line that passes through the midpoints of $BC$ and $AX$.

Let $ABCD$ be a trapezoid with $AB // CD$, $2|AB| = |CD|$ and $BD \perp BC$. Let $M$ be the midpoint of $CD$ and let $E$ be the intersection $BC$ and $AD$. Let $O$ be the intersection of $AM$ and $BD$. Let $N$ be the intersection of $OE$ and $AB$. (a) Prove that $ABMD$ is a rhombus. (b) Prove that the line $DN$ passes through the midpoint of the line segment $BE$.

Let $ABC$ be an isosceles triangle $ABC$ with base $BC$ insribed in a circle. The segment $AD$ is the diameter of the circle, and point $P$ lies on the smaller arc $BD$. Line $DP$ intersects rays $AB$ and $AC$ at points $M$ and $N$, and the lines $BP$ and $CP$ intersects the line $AD$ at points $Q$ and $R$. Prove that the midpoint of the segment $MN$ lies on the circumscribed circle of triangle $PQR$.

Let $ABC$ be an acute triangle, with $AB \ne AC$. Let $D$ be the midpoint of the line segment $BC$, and let $E$ and $F$ be the projections of $D$ onto the sides $AB$ and $AC$, respectively. If $M$ is the midpoint of the line segment $EF$, and $O$ is the circumcenter of triangle $ABC$, prove that the lines $DM$ and $AO$ are parallel. [hide=PS] As source was given [url=https://artofproblemsolving.com/community/c629086_caucasus_mathematical_olympiad]Caucasus MO[/url], but I was unable to find this problem in the contest collections [/hide]

Let $ABC$ be an acute-angled triangle, $M$ the midpoint of the side $AC$ and $F$ the foot on $AB$ of the altitude through the vertex $C$. Prove that $AM = AF$ holds if and only if $\angle BAC = 60^o$. (Karl Czakler)

Let $ABC$ be a given triangle and $\ell$ be a line that meets the lines $BC, CA$ and $AB$ in $A_1,B_1$ and $C_1$ respectively. Let $A'$ be the midpoint, of the segment connecting the projections of $A_1$ onto the lines $AB$ and $AC$. Construct, analogously the points $B'$ and $C'$. (a) Show that the points $A', B'$ and $C'$ are collinear on some line $\ell'$. (b) Show that if $\ell$ contains the circumcenter of the triangle $ABC$, then $\ell' $ contains the center of it's Euler circle.