Found problems: 533
Let $P,Q,R$ be points on the same side of a line $g$ in the plane. Let $M$ and $N$ be the feet of the perpendiculars from $P$ and $Q$ to $g$ respectively. Point $S$ lies between the lines $PM$ and $QN$ and satisfies and satisfies $PM = PS$ and $QN = QS$. The perpendicular bisectors of $SM$ and $SN$ meet in a point $R$. If the line $RS$ intersects the circumcircle of triangle $PQR$ again at $T$, prove that $S$ is the midpoint of $RT$.
Let $\Gamma_1$ and $\Gamma_2$ be two non-overlapping circles. $A,C$ are on $\Gamma_1$ and $B,D$ are on $\Gamma_2$ such that $AB$ is an external common tangent to the two circles, and $CD$ is an internal common tangent to the two circles. $AC$ and $BD$ meet at $E$. $F$ is a point on $\Gamma_1$, the tangent line to $\Gamma_1$ at $F$ meets the perpendicular bisector of $EF$ at $M$. $MG$ is a line tangent to $\Gamma_2$ at $G$. Prove that $MF=MG$.
Initially, there are three different points on the plane. Every minute, three points are chosen, for example $A$, $B$ and $C$, and a new point $D$ is generated which is symmetric to $A$ with respect to the perpendicular bisector of line segment $BC$. 24 hours later, it turns out that among all the points that were generated, there exist three collinear points. Prove that the three initial points were also collinear. (Author: V. Shmarov)
In triangle $ABC$, $\angle B=2\angle C$ and $\angle A>90^\circ$. Let $D$ be the point on the line $AB$ such that $CD$ is perpendicular to $AC$, and let $M$ be the midpoint of $BC$. Prove that $\angle AMB=\angle DMC$.
Let $ABC$ be a triangle, and $M$ and $N$ variable points on $AB$ and $AC$ respectively, such that both $M$ and $N$ do not lie on the vertices, and also, $AM \times MB = AN \times NC$. Prove that the perpendicular bisector of $MN$ passes through a fixed point.
Point $ D$ is chosen on the side $ AB$ of triangle $ ABC$. Point $ L$ inside the triangle $ ABC$ is such that $ BD=LD$ and $ \angle LAB=\angle LCA=\angle DCB$. It is known that $ \angle ALD+\angle ABC=180^\circ$. Prove that $ \angle BLC=90^\circ$.
Points $B,C$ vary on two fixed rays emanating from point $A$ such that $AB+AC$ is constant. Show that there is a point $D$, other than $A$, such that the circumcircle of triangle $ABC$ passes through $D$ for all possible choices of $B, C$.
Let the incircle $k$ of the triangle $ABC$ touch its side $BC$ at $D$. Let the line $AD$ intersect $k$ at $L \neq D$ and denote the excentre of $ABC$ opposite to $A$ by $K$. Let $M$ and $N$ be the midpoints of $BC$ and $KM$ respectively.
Prove that the points $B, C, N,$ and $L$ are concyclic.