Let $ABC$ be a triangle with circumcircle $\omega$ . Point $D$ lies on the arc $BC$ of $\omega$ and is different than $B,C$ and the midpoint of arc $BC$. Tangent of $\Gamma$ at $D$ intersects lines $BC$, $CA$, $AB$ at $A',B',C'$, respectively. Lines $BB'$ and $CC'$ intersect at $E$. Line $AA'$ intersects the circle $\omega$ again at $F$. Prove that points $D,E,F$ are collinear. (Saudi Arabia)

Circles ${{w} _ {1}}$ and ${{w} _ {2}}$ with centers ${{O} _ {1}}$ and ${{O} _ {2}}$ intersect at points $A$ and $B$, respectively. The line ${{O} _ {1}} {{O} _ {2}}$ intersects ${{w} _ {1}}$ at the point $Q$, which does not lie inside the circle ${{w} _ {2}}$, and ${{w} _ {2}}$ at the point $X$ lying inside the circle ${{w} _ {1} }$. Around the triangle ${{O} _ {1}} AX$ circumscribe a circle ${{w} _ {3}}$ intersecting the circle ${{w} _ {1}}$ for the second time in point $T$. The line $QT$ intersects the circle ${{w} _ {3}}$ at the point $K$, and the line $QB$ intersects ${{w} _ {2}}$ the second time at the point $H$. Prove that a) points $T, \, \, X, \, \, B$ lie on one line; b) points $K, \, \, X, \, \, H$ lie on one line. (Vadym Mitrofanov)

Given a square $ABCD$, point $E$ is the midpoint of $AD$. Let $F$ be the foot of the perpendicular drawn from point $B$ on $EC$. Point $K$ on $AB$ is such that $\angle DFK = 90^o$. The point $N$ on the $CE$ is such that $\angle NKB = 90^o$. Prove that the point $N$ lies on the segment $BD$. (Matvii Kurskyi) [img]https://cdn.artofproblemsolving.com/attachments/4/2/d42b8c8117ec1d5e5c5b981904779b156fce93.png[/img]

Let $P_1$ and $P_2$ be two non-parallel planes in space, and $A$ a point that does not It is in none of them. For each point $X$, let $X_1$ denote its reflection with respect to $P_1$, and $X_2$ its reflection with respect to $P_2$. Determine the locus of points $X$ for the which $X_1, X_2$ and $A$ are collinear.

Triangles $ABC$ and $A_1B_1C_1$ are similar and differently oriented. On the segment $AA_1$, a point $A'$ is taken such that $AA' / A_1A'= BC / B_1C_1$. We similarly construct $B'$ and $C'$. Prove that $A', B',C'$ lie on one straight line.

In a certain geometry we operate with two types of elements, points and lines, related to each other by the following axioms: [b]I.[/b] Given two points $A$ and $B$, there is a unique line $(AB)$ that passes through both. [b]II. [/b]There are at least two points on a line. There are three points not situated on a straight line. [b]III.[/b] When a point $B$ is located between $A$ and $C$, then $B$ is also between $C$ and $A$. ($A, B, C$ are three different points on a line.) [b]IV.[/b] Given two points $A$ and $C$, there exists at least one point $B$ on the line $(AC)$ of the form that C is between $A$ and $B$. [b]V.[/b] Among three points located on the same straight line, one at most is between the other two. [b]VI.[/b] If $A, B, C$ are three points not lying on the same line and a is a line that does not contain any of the three, when the line passes through a point on segment [AB] , then it goes through one of the $[BC]$ , or it goes through one of the [AC] . (We designate by [AB] the set of points that lie between $A$ and $B$.) From the previous axioms, prove the following propositions: Theorem 1. Between points A and C there is at least one point $B$. Theorem 2. Among three points located on a line, one is always between the two others.

On the sides $AB$ and $BC$ arbitrarily mark points $M$ and $N$, respectively. Let $P$ be the point of intersection of segments $AN$ and $BM$. In addition, we note the points $Q$ and $R$ such that quadrilaterals $MCNQ$ and $ACBR$ are parallelograms. Prove that the points $P,Q$ and $R$ lie on one line.

Let $I$ be the center of a circle inscribed in triangle $ABC$, in which $\angle BAC = 60 ^o$ and $AB \ne AC$. The points $D$ and $E$ were marked on the rays $BA$ and $CA$ so that $BD = CE = BC$. Prove that the line $DE$ passes through the point $I$.

Circle $k$ with center $O$ is touched from inside by two circles in points $S$ and $T,$ respectively. Let those two circles intersect at points $M$ and $N$, such that $N$ is closer to line $ST$. Prove that $OM$ and $MN$ are perpendicular iff $S$, $N$ and $T$ are collinear

Let $ABC$ be a triangle, $I$ its incenter, and $D$ a point on the arc $BC$ of the circumcircle of $ABC$ not containing $A$. The bisector of the angle $\angle ADB$ intesects the segment $AB$ at $E$. The bisector of the angle $\angle CDA$ intesects the segment $AC$ at $F$. Prove that the points $E, F,I$ are collinear. (Malik Talbi)

On the horizontal line from left to right are the points $P, \, \, Q, \, \, R, \, \, S$. Construct a square $ABCD$, for which on the line $AD$ lies lies the point $P$, on the line $BC$ lies the point $Q$, on the line $AB$ lies the point $R$, on the line $CD$ lies the point $S $.

$A, B, C, D$ are consecutive vertices of a regular $7$-gon. $AL$ and $AM$ are tangents to the circle center $C$ radius $CB$. $N$ is the intersection point of $AC$ and $BD$. Show that $L, M, N$ are collinear.

Let $\omega_1, \omega_2$ be two circles with different radii, and let $H$ be the exsimilicenter of the two circles. A point X outside both circles is given. The tangents from $X$ to $\omega_1$ touch $\omega_1$ at $P, Q$, and the tangents from $X$ to $\omega_2$ touch $\omega_2$ at $R, S$. If $PR$ passes through $H$ and is not a common tangent line of $\omega_1, \omega_2$, prove that $QS$ also passes through $H$.

Two congruent equilateral triangles $ABC$ and $A'B'C'$ in the plane are given. Show that the midpoints of the segments $AA',BB', CC'$ either are collinear or form an equilateral triangle.

In the convex quadrilateral of the $ABCD$, the diagonals of $AC$ and $BD$ are mutually perpendicular and intersect at point $E$. On the side of $AD$, a point $P$ is chosen such that $PE = EC$. The circumscribed circle of the triangle $BCD$ intersects the segment $AD$ at the point $Q$. The circle passing through point $A$ and tangent to the line $EP$ at point $P$ intersects the segment $AC$ at point $R$. It turns out that points $B, Q, R$ are collinear. Prove that $\angle BCD = 90^o$.

Given an arbitrary triangle $ABC$, with $\angle A = 60^o$ and $AC < AB$. A circle with diameter $BC$, intersects $AB$ and $AC$ at $F$ and $E$, respectively. Lines $BE$ and $CF$ intersect at $D$. Let $\Gamma$ be the circumcircle of $BCD$, where the center of $\Gamma$ is $O$. Circle $\Gamma$ intersects the line $AB$ and the extension of $AC$ at $M$ and $N$, respectively. $MN$ intersects $BC$ at $P$. Prove that points $A$, $P$, $O$ lie on the same line.

Given a circle $\omega$, a point $A$ lying inside $\omega$, and point $B$ ($B \ne A$). All possible triangles $BXY$ are considered, such that the points $X$ and $Y$ lie on $\omega$ and the chord $XY$ passes through the point $A$. Prove that the centers of the circumcircles of the triangles $BXY$ lie on the same straight line.

Let $ABC$ be an acute-angled and non-isosceles triangle with orthocenter $H$. Let $O$ be the center of the circumscribed circle of triangle $ABC$ and let $K$ be center of the circumscribed circle of triangle $AHO$. Prove that the reflection of $K$ wrt $OH$ lies on $BC$.

Let $ABC$ be an acute triangle. Suppose that circle $\Gamma_1$ has it's center on the side $AC$ and is tangent to the sides $AB$ and $BC$, and circle $\Gamma_2$ has it's center on the side $AB$ and is tangent to the sides $AC$ and $BC$. The circles $\Gamma_1$ and $ \Gamma_2$ intersect at two points $P$ and $Q$. Show that if $A, P, Q$ are collinear, then $AB = AC$.

Let $ABC$ be a right triangle with hypotenuse $BC$. The tangent to the circumcircle of triangle $ABC$ at $A$ intersects the line $BC$ at $T$. The points $D$ and $E$ are chosen so that $AD = BD, AE = CE,$ and $\angle CBD = \angle BCE < 90^{\circ}$. Prove that $D, E,$ and $T$ are collinear. Proposed by [i]Nikola Velov, Macedonia[/i]

Two circles $S_1$ and $S_2$ touch externally at $F$. their external common tangent touches $S_1$ at $A$ and $S_2$ at $B$. A line, parallel to $AB$ and tangent to $S_2$ at $C$, intersects $S_1$ at $D$ and $E$. Prove that points $A,F,C$ are collinear. (A. Kalinin)

On the diagonals $AC$ and $CE$ of a regular hexagon $ABCDEF$ with side $1$ we mark points$ M$ and $N$ such that $AM = CN = a$. Find $a$ if the points $B, M, N$ lie on the same line.

Let $A$ be a point on the circle $\omega$ centred at $B$ and $\Gamma$ a circle centred at $A$. For $i=1,2,3$, a chord $P_iQ_i$ of $\omega$ is tangent to $\Gamma$ at $S_i$ and another chord $P_iR_i$ of $\omega$ is perpendicular to $AB$ at $M_i$. Let $Q_iT_i$ be the other tangent from $Q_i$ to $\Gamma$ at $T_i$ and $N_i$ be the intersection of $AQ_i$ with $M_iT_i$. Prove that $N_1,N_2,N_3$ are collinear.

Two circles $w_1$ and $w_2$ intersect at points $A$ and $B$. Tangents $\ell_1$ and $\ell_2$ respectively are drawn to them through point $A$. The perpendiculars dropped from point $B$ to $\ell_2$ and $\ell_1$ intersects the circles $w_1$ and $w_2$, respectively, at points $K$ and $N$. Prove that points $K, A$ and $N$ lie on one straight line.

Let $ABCD$ be a rhombus and $P$ be a point on its side $BC$. The circle passing through $A, B$, and $P$ intersects $BD$ once more at the point $Q$ and the circle passing through $C,P$ and $Q$ intersects $BD$ once more at the point $R$. Prove that $A, R$ and $P$ lie on the one straight line. (D. Fomin, Leningrad)