In triangle $\vartriangle ABC$ holds $\angle A= 40^o$ and $\angle B = 20^o$ . The point $P$ lies on the line $AC$ such that $C$ is between $A$ and $P$ and $| CP | = | AB | - | BC |$. Calculate the $\angle CBP$.

The diagram below shows the first three figures of a sequence of figures. The first figure shows an equilateral triangle $ABC$ with side length $1$. The leading edge of the triangle going in a clockwise direction around $A$ is labeled $\overline{AB}$ and is darkened in on the figure. The second figure shows the same equilateral triangle with a square with side length $1$ attached to the leading clockwise edge of the triangle. The third figure shows the same triangle and square with a regular pentagon with side length $1$ attached to the leading clockwise edge of the square. The fourth figure in the sequence will be formed by attaching a regular hexagon with side length $1$ to the leading clockwise edge of the pentagon. The hexagon will overlap the triangle. Continue this sequence through the eighth figure. After attaching the last regular figure (a regular decagon), its leading clockwise edge will form an angle of less than $180^\circ$ with the side $\overline{AC}$ of the equilateral triangle. The degree measure of that angle can be written in the form $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] size(250); defaultpen(linewidth(0.7)+fontsize(10)); pair x[],y[],z[]; x[0]=origin; x[1]=(5,0); x[2]=rotate(60,x[0])*x[1]; draw(x[0]--x[1]--x[2]--cycle); for(int i=0;i<=2;i=i+1) { y[i]=x[i]+(15,0); } y[3]=rotate(90,y[0])*y[2]; y[4]=rotate(-90,y[2])*y[0]; draw(y[0]--y[1]--y[2]--y[0]--y[3]--y[4]--y[2]); for(int i=0;i<=4;i=i+1) { z[i]=y[i]+(15,0); } z[5]=rotate(108,z[4])*z[2]; z[6]=rotate(108,z[5])*z[4]; z[7]=rotate(108,z[6])*z[5]; draw(z[0]--z[1]--z[2]--z[0]--z[3]--z[4]--z[2]--z[7]--z[6]--z[5]--z[4]); dot(x[2]^^y[2]^^z[2],linewidth(3)); draw(x[2]--x[0]^^y[2]--y[4]^^z[2]--z[7],linewidth(1)); label("A",(x[2].x,x[2].y-.3),S); label("B",origin,S); label("C",x[1],S);[/asy]

Let $ABCD$ be a convex quadrilateral with $\angle A=60^o$. Let $E$ and $Z$ be the symmetric points of $A$ wrt $BC$ and $CD$ respectively. If the points $B,D,E$ and $Z$ are collinear, then calculate the angle $\angle BCD$.

It is known that in the triangle $ABC$ the distance from the intersection point of the angle bisector to each of the vertices of the triangle does not exceed the diameter of the circle inscribed in this triangle. Find the angles of the triangle $ABC$. (Grigory Filippovsky)

In the isosceles triangle $ABC$, $ (AB = BC)$ the bisector $AD$ was drawn, and in the triangle $ABD$ the bisector $DE$ was drawn. Find the values of the angles of the triangle $ABC$, if it is known that the bisectors of the angles $ABD$ and $AED$ intersect on the line $AD$. (Fedak Ivan)

$O$ and $I$ are the circumcentre and incentre of $\vartriangle ABC$ respectively. Suppose $O$ lies in the interior of $\vartriangle ABC$ and $I$ lies on the circle passing through $B, O$, and $C$. What is the magnitude of $\angle B AC$ in degrees?

Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.

The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$

Let ABC a triangle, let K be the foot of the bisector relative to BC and J be the foot of the trisectrix relative to BC closer to the side AC (3* m(JAC)=m(CAB) ). Let C' and B' be two point on the line AJ on the side of J with respect to A, such that AC'=AC and AB=AB'. Prove that ABB'C is cyclic if and only if lines C'K and BB' are parallel.

Let $ABCD$ be a parallelogram. A line through $C$ crosses the side $AB$ at an interior point $X$, and the line $AD$ at $Y$. The tangents of the circle $AXY$ at $X$ and $Y$, respectively, cross at $T$. Prove that the circumcircles of triangles $ABD$ and $TXY$ intersect at two points, one lying on the line $AT$ and the other one lying on the line $CT$.

Circles $\omega_1$ and $\omega_2$ are externally tangent to each other at $P$. A random line $\ell$ cuts $\omega_1$ at $A$ and $C$ and $\omega_2$ at $B$ and $D$ (points $A,C,B,D$ are in this order on $\ell$). Line $AP$ meets $\omega_2$ again at $E$ and line $BP$ meets $\omega_1$ again at $F$. Prove that the radical axis of circles $(PCD)$ and $(PEF)$ is parallel to $\ell$. \\ \\ [i](Vlad Robu)[/i]

In square $ABCD,$ points $E$ and $F$ are chosen in the interior of sides $BC$ and $CD$, respectively. The line drawn from $F$ perpendicular to $AE$ passes through the intersection point $G$ of $AE$ and diagonal $BD$. A point $K$ is chosen on $FG$ such that $|AK|= |EF|$. Find $\angle EKF.$

Let $ABC$ be an isosceles triangle with $AB = AC$. Suppose $P,Q,R$ are points on segments $AC, AB, BC$ respectively such that $AP = QB$, $\angle PBC = 90^\circ - \angle BAC$ and $RP = RQ$. Let $O_1, O_2$ be the circumcenters of $\triangle APQ$ and $\triangle CRP$. Prove that $BR = O_1O_2$. [i]Proposed by Atul Shatavart Nadig[/i]

Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$. Let $M_B$ and $M_C$ be the midpoints of $AC$ and $AB$, respectively. Place points $X$ and $Y$ on line $BC$ such that $\angle HM_BX = \angle HM_CY = 90^{\circ}$. Prove that triangles $OXY$ and $HBC$ are similar.

Let $ABC$ be a triangle with $AB = AC$ and let $D$ be the foot of the height drawn from $A$ to $BC$. Let $P$ be a point inside the triangle $ADC$ such that $\angle APB> 90^o$ and $\angle PAD + \angle PBD = \angle PCD$. The $CP$ and $AD$ lines are cut at $Q$ and the $BP$ and $AD$ lines cut into $R$. Let $T$ be a point in segment $AB$ such that $\angle TRB = \angle DQC$ and let S be a point in the extension of the segment $AP$ (on the $P$ side) such that $\angle PSR = 2 \angle PAR$. Prove that $RS = RT$.

Let $ABCD$ be a parallelogram. Point $E$ lies on segment $CD$ such that \[2\angle AEB=\angle ADB+\angle ACB,\] and point $F$ lies on segment $BC$ such that \[2\angle DFA=\angle DCA+\angle DBA.\] Let $K$ be the circumcenter of triangle $ABD$. Prove that $KE=KF$. [i]Merlijn Staps[/i]

Let $AK$ and $AT$ be the bisector and the median of an acute-angled triangle $ABC$ with $AC > AB$. The line $AT$ meets the circumcircle of $ABC$ at point $D$. Point $F$ is the reflection of $K$ about $T$. If the angles of $ABC$ are known, find the value of angle $FDA$.

Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.

Given a convex quadrilateral $ABCD$. We denote $I_A,I_B, I_C$ and $I_D$ centers of $\omega_A, \omega_B,\omega_C $and $\omega_D$,inscribed In the triangles $DAB, ABC, BCD$ and $CDA$, respectively.It turned out that $\angle BI_AA + \angle I_CI_AI_D = 180^\circ$. Prove that $\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}$. (A. Kuznetsov)

Let $ABCD$ be a trapezoid with $AD \parallel BC$ and $\angle BCD < \angle ABC < 90^\circ$. Let $E$ be the intersection point of the diagonals $AC$ and $BD$. The circumcircle $\omega$ of $\triangle BEC$ intersects the segment $CD$ at $X$. The lines $AX$ and $BC$ intersect at $Y$, while the lines $BX$ and $AD$ intersect at $Z$. Prove that the line $EZ$ is tangent to $\omega$ iff the line $BE$ is tangent to the circumcircle of $\triangle BXY$.

Points $M$ and $N$ are the midpoints of sides $AC$ and $BC$ of a triangle $ABC$. It is known that $\angle MAN = 15^o$ and $\angle BAN = 45^o$. Find the value of angle $\angle ABM$. (A. Blinkov)

Let $BB_1$ and $CC_1$ be the altitudes of the acute-angled triangle $ABC$. From the point $B_1$ the perpendiculars $B_1E$ and $B_1F$ are drawn on the sides $AB$ and $BC$ of the triangle, respectively, and from the point $C_1$ the perpendiculars $C_1 K$ and $C_1L$ on the sides $AC$ and $BC$, respectively. It turned out that the lines $EF$ and $KL$ are perpendicular. Find the measure of the angle $A$ of the triangle $ABC$. (Alexander Dunyak)

Let $ABC$ be an acute-angled triangle \( ABC \) with \( AC > BC \) and incenter \( I \). Let \( \omega \) be the mixtilinear circle at vertex \( C \), i.e. the circle internally tangent to the circumcircle of \( \triangle ABC \) and also tangent to lines \( AC \) and \( BC \). A circle \( \Gamma \) passes through points \( A \) and \( B \) and is tangent to \( \omega \) at point \( T \), with \( C \notin \Gamma \) and \( I \) being inside \( \triangle ATB \). Prove that: $$\angle CTB + \angle ATI = 180^\circ + \angle BAI - \angle ABI.$$

Figure shows a semicircle with diameter $AD$. The chords $AC$ and $BD$ meet at $P$. $Q$ is the foot of the perpendicular from $P$ to $AD$. find $\angle BCQ$ in terms of $\theta$ and $\phi$ . [img]https://cdn.artofproblemsolving.com/attachments/a/2/2781050e842b2dd01b72d246187f4ed434ff69.png[/img]

A pentagon $ABCDE$ is circumscribed about a circle. The angles at the vertices $A{}$, $C{}$ and $E{}$ of the pentagon are equal to $100^\circ$. Find the measure of the angle $\angle ACE$.