Given a triangle $ABC$ in which $|AB|>|AC|$. Let $P$ be the midpoint of the side $BC$, and $S$ the point in which the angle bisector of $\angle BAC$ intersects that side. The parallel with the line $AS$ through the point $P$ intersects lines $AB$ and $AC$ at points $X$ and $Y$ respectively . Let $Z$ be the point be such that $Y$ is the midpoint of the length $XZ$ and let the lines $BY$ and $CZ$ intersect at point $D$. Prove that the angle bisector of $\angle BDC$ is parallel to the lines $AS$.

A circle $\omega$ with center $I$ is inscribed into a segment of the disk, formed by an arc and a chord $AB$. Point $M$ is the midpoint of this arc $AB$, and point $N$ is the midpoint of the complementary arc. The tangents from $N$ touch $\omega$ in points $C$ and $D$. The opposite sidelines $AC$ and $BD$ of quadrilateral $ABCD$ meet in point $X$, and the diagonals of $ABCD$ meet in point $Y$. Prove that points $X, Y, I$ and $M$ are collinear.

Let a cyclic quadrilateral $ ABCD$, $ AC \cap BD \equal{} E$ and let a circle $ \Gamma$ internally tangent to the arch $ BC$ (that not contain $ D$) in $ T$ and tangent to $ BE$ and $ CE$. Call $ R$ the point where the angle bisector of $ \angle ABC$ meet the angle bisector of $ \angle BCD$ and $ S$ the incenter of $ BCE$. Prove that $ R$, $ S$ and $ T$ are collinear. [i](Gabriel Giorgieri)[/i]

Let $A$ and $B$ be points on the circle $k$ with center $O$, so that $AB> AO$. Let $C$ be the intersection of the bisectors of $\angle OAB$ and $k$, different from $A$. Let $D$ be the intersection of the straight line $AB$ with the circumcircle of the triangle $OBC$, different from $B$. Show that $AD = AO$ .

Consider a line segment $[AB]$ with midpoint $M$ and perpendicular bisector $m$. For each point$ X \ne M$ on m consider we are the intersection point $Y$ of the line $BX$ with the bisector from the angle $\angle BAX$. As $X$ approaches $M$, then approaches $Y$ to a point of $[AB]$. Which? [img]https://cdn.artofproblemsolving.com/attachments/a/3/17d72a23011a9ec22deb20184717cc9c020a2b.png[/img] [hide=original wording]Beschouw een lijnstuk [AB] met midden M en middelloodlijn m. Voor elk punt X 6= M op m beschouwenwe het snijpunt Y van de rechte BX met de bissectrice van de hoek < BAX . Als X tot M nadert, dan nadert Y tot een punt van [AB]. Welk? [/hide]

In a triangle $ABC$ the bisectors $BB'$ and $CC'$ are drawn. After that, the whole picture except the points $A, B'$, and $C'$ is erased. Restore the triangle using a compass and a ruler. (A.Karlyuchenko)

Triangle $ABC$ has $AC = 3$, $BC = 5$, $AB = 7$. A circle is drawn internally tangent to the circumcircle of $ABC$ at $C$, and tangent to $AB$. Let $D$ be its point of tangency with $AB$. Find $BD - DA$. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(6cm); real labelscalefactor = 2.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.5, xmax = 7.01, ymin = -3, ymax = 8.02; /* image dimensions */ /* draw figures */ draw(circle((1.37,2.54), 5.17)); draw((-2.62,-0.76)--(-3.53,4.2)); draw((-3.53,4.2)--(5.6,-0.44)); draw((5.6,-0.44)--(-2.62,-0.76)); draw(circle((-0.9,0.48), 2.12)); /* dots and labels */ dot((-2.62,-0.76),dotstyle); label("$C$", (-2.46,-0.51), SW * labelscalefactor); dot((-3.53,4.2),dotstyle); label("$A$", (-3.36,4.46), NW * labelscalefactor); dot((5.6,-0.44),dotstyle); label("$B$", (5.77,-0.17), SE * labelscalefactor); dot((0.08,2.37),dotstyle); label("$D$", (0.24,2.61), SW * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("$7$",(-3.36,4.46)--(5.77,-0.17), NE * labelscalefactor); label("$3$",(-3.36,4.46)--(-2.46,-0.51),SW * labelscalefactor); label("$5$",(-2.46,-0.51)--(5.77,-0.17), SE * labelscalefactor); /* end of picture */ [/asy]

Let D be an arbitrary point inside $\Delta ABC$. Let $\Gamma$ be the circumcircle of $\Delta BCD$. The external angle bisector of $\angle ABC$ meets $\Gamma$ again at $E$. The external angle bisector of $\angle ACB$ meets $\Gamma$ again at $F$. The line $EF$ meets the extension of $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the circumcircles of $\Delta BFP$ and $\Delta CEQ$ always pass through the same fixed point regardless of the position of $D$. (Assume all the labelled points are distinct.)

Let $D$ be a point on side $[AB]$ of triangle $ABC$ with $|AB|=|AC|$ such that $[CD]$ is an angle bisector and $m(\widehat{ABC})=40^\circ$. Let $F$ be a point on the extension of $[AB]$ after $B$ such that $|BC|=|AF|$. Let $E$ be the midpoint of $[CF]$. If $G$ is the intersection of lines $ED$ and $AC$, what is $m(\widehat{FBG})$? $ \textbf{(A)}\ 150^\circ \qquad\textbf{(B)}\ 135^\circ \qquad\textbf{(C)}\ 120^\circ \qquad\textbf{(D)}\ 105^\circ \qquad\textbf{(E)}\ \text{None of above} $

Let $ABC$ be a triangle and $D$ be a point on $BC$ such that $AB+BD=AC+CD$. The line $AD$ intersects the incircle of triangle $ABC$ at $X$ and $Y$ where $X$ is closer to $A$ than $Y$ i. Suppose $BC$ is tangent to the incircle at $E$, prove that: 1) $EY$ is perpendicular to $AD$; 2) $XD=2IM$ where $I$ is the incentre and $M$ is the midpoint of $BC$.

Given triangle $ ABC$ of area 1. Let $ BM$ be the perpendicular from $ B$ to the bisector of angle $ C$. Determine the area of triangle $ AMC$.

The circles $C_1$ and $C_2$ with radii $r_1$ and $r_2$ touch the line $p$ at the point $P$. $C_1$ lies entirely within $C_2$. Line $q \perp p$ intersects $p$ at $S$ and touches $C_1$ at $R$. $q$ intersects $C_2$ at $M$ and $N$, where $N$ is between $R$ and $S$. a) Prove that line $PR$ bisects angle $\angle MPN$. b) Calculate the ratio $r_1 : r_2$ if line $PN$ bisects angle $\angle RPS$.

Points $A$, $B$, $C$ and $D$ are chosen on a line in that order, with $AB$ and $CD$ greater than $BC$. Equilateral triangles $APB$, $BCQ$ and $CDR$ are constructed so that $P$, $Q$ and $R$ are on the same side with respect to $AD$. If $\angle PQR=120^\circ$, show that \[\frac{1}{AB}+\frac{1}{CD}=\frac{1}{BC}.\]

In an acute scalene triangle $ABC$ the bisector of the acute angle between the altitudes $AA_1$ and $CC_1$ meets the sides $AB$ and $BC$ at $P$ and $Q$ respectively. The bisector of the angle $B$ intersects the segment joining the orthocenter of $ABC$ and the midpoint of $AC$ at point $R$. Prove that $P$, $B$, $Q$, $R$ lie on a circle.

In $\vartriangle ABC, \angle A = 66^o$ and $| AB | <| AC |$. The outer bisector in $A$ intersects $BC$ in $D$ and $| BD | = | AB | + | AC |$. Determine the angles of $\vartriangle ABC$.

Denote $\overline{w_a}, \overline{w_b}, \overline{w_c}$ the external angle-bisectors in triangle $ABC$, prove that $$\sum \limits_{cyc} \frac{1}{w_a}\leq \sqrt{\frac{(s^2 - r^2 - 4Rr)(8R^2 - s^2 - r^2 - 2Rr)}{8s^2R^2r}}$$

In an acute triangle $ABC$, altitudes at vertices $A$ and $B$ and bisector line at angle $C$ intersect the circumcircle again at points $A_1, B_1$ and $C_0$. Using the straightedge and compass, reconstruct the triangle by points $A_1, B_1$ and $C_0$.

Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD$. The two diagonals meet at $P$. Let $O$ be the foot of the perpendicular from $P$ to $EF$. Show that $\angle BOC=\angle AOD$.

Given a circumcircle $(O)$ and two fixed points $B,C$ on $(O)$. $BC$ is not the diameter of $(O)$. A point $A$ varies on $(O)$ such that $ABC$ is an acute triangle. $E,F$ is the foot of the altitude from $B,C$ respectively of $ABC$. $(I)$ is a variable circumcircle going through $E$ and $F$ with center $I$. a) Assume that $(I)$ touches $BC$ at $D$. Probe that $\frac{DB}{DC}=\sqrt{\frac{\cot B}{\cot C}}$. b) Assume $(I)$ intersects $BC$ at $M$ and $N$. Let $H$ be the orthocenter and $P,Q$ be the intersections of $(I)$ and $(HBC)$. The circumcircle $(K)$ going through $P,Q$ and touches $(O)$ at $T$ ($T$ is on the same side with $A$ wrt $PQ$). Prove that the interior angle bisector of $\angle{MTN}$ passes through a fixed point.

Let $\omega,\omega_1,\omega_2$ be three mutually tangent circles such that $\omega_1,\omega_2$ are externally tangent at $P$, $\omega_1,\omega$ are internally tangent at $A$, and $\omega,\omega_2$ are internally tangent at $B$. Let $O,O_1,O_2$ be the centers of $\omega,\omega_1,\omega_2$, respectively. Given that $X$ is the foot of the perpendicular from $P$ to $AB$, prove that $\angle{O_1XP}=\angle{O_2XP}$. [i]David Yang.[/i]

Let $ABC$ be an acute-angled triangle with circumference $\Omega$. Let the angle bisectors of $\angle B$ and $\angle C$ intersect $\Omega$ again at $M$ and $N$. Let $I$ be the intersection point of these angle bisectors. Let $M'$ and $N'$ be the respective reflections of $M$ and $N$ in $AC$ and $AB$. Prove that the center of the circle passing through $I$, $M'$, $N'$ lies on the altitude of triangle $ABC$ from $A$. [i]Proposed by Victor Domínguez and Ariel García[/i]

Excircle of $ABC$ is tangent to the side $BC$ at point $K$ and is tangent to the extension of $AB$ at point $L$. Another excircle is tangent to extensions of sides $AB$ and $BC$ at points $M$ and $N$. Lines $KL$ and $MN$ intersect at point $X$. Prove that $CX$ is the bisector of angle $ACN$. [I]Proposed by S. Berlov[/i]

Let $A_{1}$, $B_{1}$, $C_{1}$ be the feet of the altitudes of an acute-angled triangle $ABC$ issuing from the vertices $A$, $B$, $C$, respectively. Let $K$ and $M$ be points on the segments $A_{1}C_{1}$ and $B_{1}C_{1}$, respectively, such that $\measuredangle KAM = \measuredangle A_{1}AC$. Prove that the line $AK$ is the angle bisector of the angle $C_{1}KM$.

Given a triangle $ \,ABC,\,$ let $ \,I\,$ be the center of its inscribed circle. The internal bisectors of the angles $ \,A,B,C\,$ meet the opposite sides in $ \,A^{\prime },B^{\prime },C^{\prime }\,$ respectively. Prove that \[ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}. \]

In an acute triangle $ABC$ the angle $\angle C$ is greater than $\angle A$. Let $E$ be such that $AE$ is a diameter of the circumscribed circle $\Gamma$ of \vartriangle ABC. Let $K$ be the intersection of $AC$ and the tangent line at $B$ to $\Gamma$. Let $L$ be the orthogonal projection of $K$ on $AE$ and let $D$ be the intersection of $KL$ and $AB$. Prove that $CE$ is the bisector of $\angle BCD$.