Points $D$ and $E$ are taken on sides $BC$ and $CA$ of a triangle $ BD\equal{}AE$. Segments $AD$ and $BE$ meet at $P$. The bisector of $\angle ACB$ intersects $AD$ and $BE$ at $Q$ and $R$ respectively. Prove that $ \frac{PQ}{PR}\equal{}\frac{AD}{BE}$.

Let $A, B,C$ be colinear points in this order, $\omega$ an arbitrary circle passing through $B$ and $C$, and $l$ an arbitrary line different from $BC$, passing through A and intersecting $\omega$ at $M$ and $N$. The bisectors of the angles $\angle CMB$ and $\angle CNB$ intersect $BC$ at $P$ and $Q$. Prove that $AP\cdot AQ = AB \cdot AC$.

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 $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.

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]

The perpendicular bisectors of the sides AB and BC of a triangle ABC meet the lines BC and AB at the points X and Z, respectively. The angle bisectors of the angles XAC and ZCA intersect at a point B'. Similarly, define two points C' and A'. Prove that the points A', B', C' lie on one line through the incenter I of triangle ABC. [i]Extension:[/i] Prove that the points A', B', C' lie on the line OI, where O is the circumcenter and I is the incenter of triangle ABC. Darij

By straightedge and compass, reconstruct a right triangle $ABC$ ($\angle C = 90^o$), given the vertices $A, C$ and a point on the bisector of angle $B$.

Triangle $ABC$ is inscribed in a circle with center $O'$. A circle with center $O$ is inscribed in triangle $ABC$. $AO$ is drawn, and extended to intersect the larger circle in $D$. Then, we must have: $\text{(A)}\ CD=BD=O'D \qquad \text{(B)}\ AO=CO=OD \qquad \text{(C)}\ CD=CO=BD \qquad\\ \text{(D)}\ CD=OD=BD \qquad \text{(E)}\ O'B=O'C=OD $ [asy] size(200); defaultpen(linewidth(0.8)+fontsize(12pt)); pair A=origin,B=(15,0),C=(5,9),O=incenter(A,B,C),Op=circumcenter(A,B,C); path incirc = incircle(A,B,C),circumcirc = circumcircle(A,B,C),line=A--3*O; pair D[]=intersectionpoints(circumcirc,line); draw(A--B--C--A--D[0]^^incirc^^circumcirc); dot(O^^Op,linewidth(4)); label("$A$",A,dir(185)); label("$B$",B,dir(355)); label("$C$",C,dir(95)); label("$D$",D[0],dir(O--D[0])); label("$O$",O,NW); label("$O'$",Op,E);[/asy]

Let $ABC$ be a triangle and the points $M$ and $N$ on the sides $AB$ respectively $BC$, such that $2 \cdot \frac{CN}{BC} = \frac{AM}{AB}$. Let $P$ be a point on the line $AC$. Prove that the lines $MN$ and $NP$ are perpendicular if and only if $PN$ is the interior angle bisector of $\angle MPC$.

Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.

Given triangle $ABC$ with $AB<AC$. The line passing through $B$ and parallel to $AC$ meets the external angle bisector of $\angle BAC$ at $D$. The line passing through $C$ and parallel to $AB$ meets this bisector at $E$. Point $F$ lies on the side $AC$ and satisfies the equality $FC=AB$. Prove that $DF=FE$.

Given is an acute triangle $ABC$ $\left(AB<AC<BC\right)$,inscribed in circle $c(O,R)$.The perpendicular bisector of the angle bisector $AD$ $\left(D\in BC\right)$ intersects $c$ at $K,L$ ($K$ lies on the small arc $\overarc{AB}$).The circle $c_1(K,KA)$ intersects $c$ at $T$ and the circle $c_2(L,LA)$ intersects $c$ at $S$.Prove that $\angle{BAT}=\angle{CAS}$. [hide=Diagram][asy]import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.94236331697463, xmax = 15.849400903703716, ymin = -5.002235438802758, ymax = 7.893104843949444; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen qqqqtt = rgb(0.,0.,0.2); draw((1.8318261909633622,3.572783369254345)--(0.,0.)--(6.,0.)--cycle, aqaqaq); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-117.14497824050169,-101.88970202103212)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-55.85706977865775,-40.60179355918817)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); /* draw figures */ draw((1.8318261909633622,3.572783369254345)--(0.,0.), uququq); draw((0.,0.)--(6.,0.), uququq); draw((6.,0.)--(1.8318261909633622,3.572783369254345), uququq); draw(circle((3.,0.7178452373968209), 3.0846882800136055)); draw((2.5345020274407277,0.)--(1.8318261909633622,3.572783369254345)); draw(circle((-0.01850947366601585,1.3533783539547308), 2.889550258039566)); draw(circle((5.553011501106743,2.4491551634556963), 3.887127532933951)); draw((-0.01850947366601585,1.3533783539547308)--(5.553011501106743,2.4491551634556963), linetype("2 2")); draw((1.8318261909633622,3.572783369254345)--(0.7798408954511686,-1.423695174396108)); draw((1.8318261909633622,3.572783369254345)--(5.22015910454883,-1.4236951743961088)); /* dots and labels */ dot((1.8318261909633622,3.572783369254345),linewidth(3.pt) + dotstyle); label("$A$", (1.5831274347452782,3.951671933606579), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.6,0.05), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.188606107156787,0.07450151636712989), NE * labelscalefactor); dot((2.5345020274407277,0.),linewidth(3.pt) + dotstyle); label("$D$", (2.3,-0.7), NE * labelscalefactor); dot((-0.01850947366601585,1.3533783539547308),linewidth(3.pt) + dotstyle); label("$K$", (-0.3447473583572136,1.6382221818835927), NE * labelscalefactor); dot((5.553011501106743,2.4491551634556963),linewidth(3.pt) + dotstyle); label("$L$", (5.631664500260511,2.580738747400365), NE * labelscalefactor); dot((0.7798408954511686,-1.423695174396108),linewidth(3.pt) + dotstyle); label("$T$", (0.5977692071595602,-1.960477431907719), NE * labelscalefactor); dot((5.22015910454883,-1.4236951743961088),linewidth(3.pt) + dotstyle); label("$S$", (5.160406217502124,-1.8747941077698307), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

In $\triangle ABC$ we have that $CC_{1}$ is an angle bisector. The points $P\in C_{1}B$, $Q\in BC$, $R\in AC$, $S\in AC_{1}$ satisfy $C_{1}P=PQ=QC$ and $CR=RS=SC_{1}$. Prove that $CC_{1}$ bisects $\angle SCP$.

Draw a line passing through a point $M$ on the angle bisector of the angle $\angle AOB$, that intersects $OA$ and $OB$ at points $K$ and $L$ respectively. Prove that the valus of the sum $\frac{1}{|OK|}+\frac{1}{|OL|}$ does not depend on the choice of the straight line passing through $M$, i.e. is defined by the size of the angle AOB and the selection of the point $M$ only.

Let $ ABC$ be a triangle in which $ AB \equal{} AC$ and let $ I$ be its in-centre. Suppose $ BC \equal{} AB \plus{} AI$. Find $ \angle{BAC}$

Let $\Delta ABC$ be a triangle with an inscribed circle centered at $I$. The line perpendicular to $AI$ at $I$ intersects $\odot (ABC)$ at $P,Q$ such that, $P$ lies closer to $B$ than $C$. Let $\odot (BIP) \cap \odot (CIQ) =S$. Prove that, $SI$ is the angle bisector of $\angle PSQ$

In $\triangle ABC$ the median $AM$ is drawn. The foot of perpendicular from $B$ to the angle bisector of $\angle BMA$ is $B_1$ and the foot of perpendicular from $C$ to the angle bisector of $\angle AMC$ is $C_1.$ Let $MA$ and $B_1C_1$ intersect at $A_1.$ Find $\frac{B_1A_1}{A_1C_1}.$

P and Q are 2 points in the area bounded by 2 rays, e and f, coming out from a point O. Describe how to construct, with a ruler and a compass only, an isosceles triangle ABC, such that his base AB is on the ray e, the point C is on the ray f, P is on AC, and Q on BC.

The diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$ intersect at point $ M$. The bisector of $ \angle ACD$ meets the ray $ BA$ at $ K$. Given that $ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$, prove that $ \angle BKC \equal{} \angle CDB$.

The angle bisectors at $A$ and $C$ in a non-isosceles triangle $ABC$ with incenter $I$ intersect its circumcircle $k$ at $A_0$ and $C_0$, respectively. The line through $I$, parallel to $AC$, intersects $A_0C_0$ at $P$. Prove that $PB$ is tangent to $k$.

$ AB$ is a fixed diameter of a circle whose center is $ O$. From $ C$, any point on the circle, a chord $ CD$ is drawn perpendicular to $ AB$. Then, as $ C$ moves over a semicircle, the bisector of angle $ OCD$ cuts the circle in a point that always: $ \textbf{(A)}\ \text{bisects the arc } AB \qquad\textbf{(B)}\ \text{trisects the arc } AB \qquad\textbf{(C)}\ \text{varies}$ $ \textbf{(D)}\ \text{is as far from }AB \text{ as from } D \qquad\textbf{(E)}\ \text{is equidistant from }B \text{ and } C$

In the triangle $ABC$ the side $AC = \tfrac {1} {2} (AB + BC) $, $BL$ is the bisector $\angle ABC$, $K, \, \, M $ - the midpoints of the sides $AB$ and $BC$, respectively. Find the value $\angle KLM$ if $\angle ABC = \beta$

Let $ ABC$ be a triangle, and let the angle bisectors of its angles $ CAB$ and $ ABC$ meet the sides $ BC$ and $ CA$ at the points $ D$ and $ F$, respectively. The lines $ AD$ and $ BF$ meet the line through the point $ C$ parallel to $ AB$ at the points $ E$ and $ G$ respectively, and we have $ FG \equal{} DE$. Prove that $ CA \equal{} CB$. [i]Original formulation:[/i] Let $ ABC$ be a triangle and $ L$ the line through $ C$ parallel to the side $ AB.$ Let the internal bisector of the angle at $ A$ meet the side $ BC$ at $ D$ and the line $ L$ at $ E$ and let the internal bisector of the angle at $ B$ meet the side $ AC$ at $ F$ and the line $ L$ at $ G.$ If $ GF \equal{} DE,$ prove that $ AC \equal{} BC.$

Let $ABCD$ be a convex quadrilateral, the incenters of $\triangle ABC$ and $\triangle ADC$ are $I,J$, respectively. It is known that $AC,BD,IJ$ concurrent at a point $P$. The line perpendicular to $BD$ through $P$ intersects with the outer angle bisector of $\angle BAD$ and the outer angle bisector $\angle BCD$ at $E,F$, respectively. Show that $PE=PF$.

A circle which is tangent to sides $AB$ and $BC$ of triangle $ABC$ is also tangent to its circumcircle at point $T$. If $I$ in the incenter of triangle $ABC$, show that $\angle ATI=\angle CTI$.