The diagonals of trapezoid $ABCD$ ($BC\parallel AD$) meet at point $O$. Points $M$ and $N$ lie on the segments $BC$ and $AD$ respectively. The tangent to the circle $AMC$ at $C$ meets the ray $NB$ at point $P$; the tangent to the circle $BND$ at $D$ meets the ray $MA$ at point $R$. Prove that $\angle BOP =\angle AOR$.

Given two circles $\omega_1$ and $\omega_2$ where $\omega_2$ is inside $\omega_1$. Show that there exists a point $P$ such that for any line $\ell$ not passing through $P$, if $\ell$ intersects circle $\omega_1$ at $A,B$ and $\ell$ intersects circle $\omega_2$ at $C,D$, where $A,C,D,B$ lie on $\ell$ in this order, then $\angle APC=\angle BPD$.

Let $AD$ be the bisector of angle $A$ of the triangle $ABC$. One considers the points M, N on the half-lines $(AB$ and $(AC$, respectively, such that $\angle MDA = \angle B$ and $\angle NDA = \angle C$. Let $AD \cap MN=\{P\}$. Prove that: $$AD^3 = AB \cdot AC\cdot AP$$

Let $M$ be the point of intersection of diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let $K$ be the point of intersection of the extension of side $AB$ (beyond$A$) with the bisector of the angle $ACD$. Let $L$ be the intersection of $KC$ and $BD$. If $MA \cdot CD = MB \cdot LD$, prove that the angle $BKC$ is equal to the angle $CDB$.

Inside the segment $AC$, an arbitrary point $B$ is selected and circles with diameters $AB$ and $BC$ are constructed. Points $M$ and $L$ are chosen on the circles (in one half-plane with respect to $AC$), respectively, so that $\angle MBA = \angle LBC$. Points $K$ and $F$ are marked, respectively, on rays $BM$ and $BL$ so that $BK = BC$ and $BF = AB$. Prove that points $M, K, F$ and $L$ lie on the same circle.

In $\triangle ABC$, points $P,Q$ satisfy $\angle PBC = \angle QBA$ and $\angle PCB = \angle QCA$, $D$ is a point on $BC$ such that $\angle PDB=\angle QDC$. Let $X,Y$ be the reflections of $A$ with respect to lines $BP$ and $CP$, respectively. Prove that $DX=DY$. [img]https://cdn.artofproblemsolving.com/attachments/a/7/f208f1651afc0fef9eef4c68ba36bf77556058.jpg[/img]

Let $ABC$ be an acute and scalene of circumcircle $\Gamma$ and orthocenter $H$. Let $A_1,B_1,C_1$ be the second intersection points of the lines $AH, BH, CH$ with $\Gamma$, respectively. The lines that pass through $A_1,B_1,C_1$ and are parallel to $BC,CA, AB$ intersect again to $\Gamma$ at $A_2,B_2,C_2$, respectively. Let $M$ be the intersection point of $AC_2$ and $BC_1, N$ the intersection point of $BA_2$ and $CA_1$, and $P$ the intersection point of $CB_2$ and $AB_1$. Prove that $\angle MNB = \angle AMP$ .

Inside the angle $AOD$, the rays $OB$ and $OC$ are drawn such that $\angle AOB = \angle COD.$ Two circles are inscribed inside the angles $\angle AOB$ and $\angle COD$ . Prove that the intersection point of the common internal tangents of these circles lies on the bisector of the angle $AOD$.

Let $ABC$ be an isosceles triangle with $AB = AC$ and let $n$ be a natural number, $n>1$. On the side $AB$ we consider the point $M$ such that $n \cdot AM = AB$. On the side $BC$ we consider the points $P_1, P_2, ....., P_ {n-1}$ such that $BP_1 = P_1P_2 = .... = P_ {n-1} C = \frac{1}{n} BC$. Show that: $\angle {MP_1A} + \angle {MP_2A} + .... + \angle {MP_ {n-1} A} = \frac{1} {2} \angle {BAC}$.

Given an isosceles triangle $ABC$ with base $AB$. Point $K$ is taken on the extension of the side $AC$ (beyond the point $C$ ) so that $\angle KBC = \angle ABC$. Denote $S$ the intersection point of angle - bisectors of $\angle BKC$ and $\angle ACB$. Lines $AB$ and $KS$ intersect at point $L$, lines $BS$ and $CL$ intersect at point $M$ . Prove that line $KM$ passes through the midpoint of the segment $BC$.

Prove that there is only one triangle whose sides are consecutive natural numbers and one of the angles is twice the other angle.

Diagonals of an isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ are perpendicular. Let $DE$ be the perpendicular from $D$ to $AB$, and let $CF$ be the perpendicular from $C$ to $DE$. Prove that angle $DBF$ is equal to half of angle $FCD$.

Given is a parallelogram $ABCD$ with $\angle A < 90^o$ and $|AB| < |BC|$. The angular bisector of angle $A$ intersects side $BC$ in $M$ and intersects the extension of $DC$ in $N$. Point $O$ is the centre of the circle through $M, C$, and $N$. Prove that $\angle OBC = \angle ODC$. [asy] unitsize (1.2 cm); pair A, B, C, D, M, N, O; A = (0,0); B = (2,0); D = (1,3); C = B + D - A; M = extension(A, incenter(A,B,D), B, C); N = extension(A, incenter(A,B,D), D, C); O = circumcenter(C,M,N); draw(D--A--B--C); draw(interp(D,N,-0.1)--interp(D,N,1.1)); draw(A--interp(A,N,1.1)); draw(circumcircle(M,C,N)); label("$\circ$", A + (0.45,0.15)); label("$\circ$", A + (0.25,0.35)); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$M$", M, SE); dot("$N$", N, dir(90)); dot("$O$", O, SE); [/asy]

Given a triangle $ABC$, the line passing through the vertex $A$ symmetric to the median $AM$ wrt the line containing the bisector of the angle $\angle BAC$ intersects the circle circumscribed around the triangle $ABC$ at points $A$ and $K$. Let $L$ be the midpoint of the segment $AK$. Prove that $\angle BLC=2\angle BAC$.

The point P lies on the side $CD$ of the parallelogram $ABCD$ with $\angle DBA = \angle CBP$. Point $O$ is the center of the circle passing through the points $D$ and $P$ and tangent to the straight line $AD$ at point $D$. Prove that $AO = OC$.

In the square $ABCD$ on the sides $AD$ and $DC$, the points $M$ and $N$ are selected so that $\angle BMA = \angle NMD = 60 { } ^ \circ $. Find the value of the angle $MBN$.