On the sides $BC$ and $CD$ of the square $ABCD$, the points $M$ and $N$ are selected in such a way that $\angle MAN= 45^o$. Using the segment $MN$, as the diameter, we constructed a circle $w$, which intersects the segments $AM$ and $AN$ at points $P$ and $Q$, respectively. Prove that the points $B, P$ and $Q$ lie on the same line.

The inscribed circle $\Omega$ of triangle $ABC$ touches the sides $AB$ and $AC$ at points $K$ and $ L$, respectively. The line $BL$ intersects the circle $\Omega$ for the second time at the point $M$. The circle $\omega$ passes through the point $M$ and is tangent to the lines $AB$ and $BC$ at the points $P$ and $Q$, respectively. Let $N$ be the second intersection point of circles $\omega$ and $\Omega$, which is different from $M$. Prove that if $KM \parallel AC$ then the points $P, N$ and $L$ lie on one line.

Let $ABC$ be an acute triangle and $\Gamma$ be its circumcircle. Line $\ell$ is tangent to $\Gamma$ at $A$ and let $D$ and $E$ be distinct points on $\ell$ such that $AD = AE$. Suppose that $B$ and $D$ lie on the same side of line $AC$. The circumcircle $\Omega_1$ of $\vartriangle ABD$ meets $AC$ again at $F$. The circumcircle $\Omega_2$ of $\vartriangle ACE$ meets $AB$ again at $G$. The common chord of $\Omega_1$ and $\Omega_2$ meets $\Gamma$ again at $H$. Let $K$ be the reflection of $H$ across line $BC$ and let $L$ be the intersection of $BF$ and $CG$. Prove that $A, K$ and $L$ are collinear.

Given trapezoid $ABCD$ with bases $AB \parallel CD$ ($AB < CD$). Let $O$ be the intersection of $AC$ and $BD$. Two straight lines from $D$ and $C$ are perpendicular to $AC$ and $BD$ intersect at $E$ , i.e. $CE \perp BD$ and $DE \perp AC$ . By analogy, $AF \perp BD$ and $BF \perp AC$ . Are three points $E , O, F$ located on the same line?

In triangle $ABC$, the points $M$ and $N$ are the midpoints of the sides $AB$ and $AC$, respectively. The bisectors of interior angles $\angle ABC$ and $\angle BCA$ intersect the line $MN$ at points $P$ and $Q$, respectively. Let $p$ be the tangent to the circumscribed circle of the triangle $AMP$ passing through point $P$, and $q$ be the tangent to the circumscribed circle of the triangle $ANQ$ passing through point $Q$. Prove that the lines $p$ and $q$ intersect on line $BC$.

Let $\vartriangle ABC$ be an acute triangle with circumcircle $\Omega$. A line passing through $A$ perpendicular to $BC$ meets $\Omega$ again at $D$. Draw two circles $\omega_b$, $\omega_c$ with $B, C$ as centers and $BD$, $CD$ as radii, respectively, and they intersect $AB$, $AC$ at $E, F,$ respectively. Let $K\ne A$ be the second intersection of $(AEF)$ and $\Omega$, and let $\omega_b$, $\omega_c$ intersect $KB$, $KC$ at $P, Q$, respectively. The circumcenter of triangle $DP Q$ is $O$, prove that $K, O, D$ are collinear. [i]proposed by Li4[/i]

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.

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]

Given triangle $ABC$. Suppose $P$ and $P_1$ are points on $BC, Q$ lies on $CA, R$ lies on $AB$, such that $$\frac{AR}{RB}=\frac{BP}{PC}=\frac{CQ}{QA}=\frac{CP_1}{P_1B}$$ Let $G$ be the centroid of triangle $ABC$ and $K = AP_1 \cap RQ$. Prove that points $P,G$, and $K$ are collinear.

Given two circles $(C_1)$ and $(C_2)$ with centers $\displaystyle{O_1}$ and $O_2$ respectively, intersecting at points $A$ and $B$. Let $AC$ και $AD$ be the diameters of $(C_1)$ and $(C_2)$ respectively . Tangent line of circle $(C_1)$ at point $A$ intersects $(C_2)$ at point $M$ and tangent line of circle $(C_2)$ at point A intersects $(C_1)$ at point $N$. Let $P$ be a point on line $AB$ such that $AB=BP$. Prove that: a) Points $B,C,D$ are collinear. b) Quadrilateral $AMPN$ is cyclic.

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 $AB = AC$.

On triangle $ABC$ the $AM$ ($M\in BC$) is median and $BB_1$ and $CC_1$ ($B_1 \in AC,C_1 \in AB$) are altitudes. The stright line $d$ is perpendicular to $AM$ at the point $A$ and intersect the lines $BB_1$ and $CC_1$ at the points $E$ and $F$ respectively. Let denoted with $\omega$ the circle passing through the points $E, M$ and $F$ and with $\omega_1$ and with $\omega_2$ the circles that are tangent to segment $EF$ and with $\omega$ at the arc $EF$ which is not contain the point $M$. If the points $P$ and $Q$ are intersections points for $\omega_1$ and $\omega_2$ then prove that the points $P, Q$ and $M$ are collinear.

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)

Let $(O_1), (O_2)$ be given two circles intersecting at $A$ and $B$. The tangent lines of $(O_1)$ at $A, B$ intersect at $O$. Let $I$ be a point on the circle $(O_1)$ but outside the circle $(O_2)$. The lines $IA, IB$ intersect circle $(O_2)$ at $C, D$. Denote by $M$ the midpoint of $C D$. Prove that $I, M, O$ are collinear.

$ABCD$ has $AB = CD$, but $AB$ not parallel to $CD$, and $AD$ parallel to $BC$. The triangle is $ABC$ is rotated about $C$ to $A'B'C$. Show that the midpoints of $BC, B'C$ and $A'D$ are collinear.

Let $I$ be the centre of the inscribed circle of the non-isosceles triangle $ABC$, and let the circle touch the sides $BC, CA, AB$ at the points $A_1, B_1, C_1$ respectively. Prove that the centres of the circumcircles of $\vartriangle AIA_1,\vartriangle BIB_1$ and $\vartriangle CIC_1$ are collinear.

Let $ABC$ be an isosceles triangle with base $AB$. We choose a point $P$ inside the triangle on altitude through $C$. The circle with diameter $CP$ intersects the straight line through $B$ and $P$ again at the point $D_P$ and the Straight through $A$ and $C$ one more time at point $E_P$. Prove that there is a point $F$ such that for any choice of $P$ the points $D_P , E_P$ and $F$ lie on a straight line. [i](Walther Janous)[/i]

Let $AD$ be the bisector of an acute-angled triangle $ABC$. The circle circumscribed around the triangle $ABD$ intersects the straight line perpendicular to $AD$ that passes through point $B$, at point $E$. Point $O$ is the center of the circumscribed circle of triangle $ABC$. Prove that the points $A, O, E$ lie on the same line.

The inscribed circle of the non-isosceles triangle $ABC$ touches sides $AB, BC$ and $AC$ at points $C_1, A_1$ and $B_1$, respectively. The circumscribed circle of the triangle $A_1BC_1$ intersects the lines $B_1A_1$ and $B_1C_1$ at the points $A_0$ and $C_0$, respectively. Prove that the orthocenter of triangle $A_0BC_0$, the center of the inscribed circle of triangle $ABC$ and the midpoint of the $AC$ lie on one straight line.

Let $ABC$ be a triangle inscribed in circle $\omega$, and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$, and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$. Prove that $O_1$, $O_2$, and the nine-point center of $ABC$ are collinear. [i]Proposed by Michael Kural[/i]

In triangle $ABC$, point $I$ is the center of the inscribed circle. $AT$ is a segment tangent to the circle circumscribed around the triangle $BIC$ . On the ray $AB$ beyond the point$ B$ and on the ray $AC$ beyond the point $C$, we draw the segments $BD$ and $CE$, respectively, such that $BD = CE = AT$. Let the point $F$ be such that $ABFC$ is a parallelogram. Prove that points $D, E$ and $F$ lie on the same line. (Dmitry Prokopenko)

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 $I$.

$3n$ points are given ($n\ge 1$) in the plane, each $3$ of them are not collinear. Prove that there are $n$ distinct triangles with the vertices those points.

Triangle $ABC$ is inscribed in circle $\omega$. Point $P$ lies inside triangle $ABC$.Lines $AP,BP$ and $CP$ intersect $\omega$ again at points $A_1$, $B_1$ and $C_1$ (other than $A, B, C$), respectively. The tangent lines to $\omega$ at $A_1$ and $B_1$ intersect at $C_2$.The tangent lines to $\omega$ at $B_1$ and $C_1$ intersect at $A_2$. The tangent lines to $\omega$ at $C_1$ and $A_1$ intersect at $B_2$. Prove that the lines $AA_2,BB_2$ and $CC_2$ are concurrent.

In the triangle $ABC$, $\angle A=90^\circ$, the bisector of $\angle B$ meets the altitude $AD$ at the point $E$, and the bisector of $\angle CAD$ meets the side $CD$ at $F$. The line through $F$ perpendicular to $BC$ intersects $AC$ at $G$. Prove that $B,E,G$ are collinear.