Let $ABC$ be a triangle such that $AB = 7$, and let the angle bisector of $\angle BAC$ intersect line $BC$ at $D$. If there exist points $E$ and $F$ on sides $AC$ and $BC$, respectively, such that lines $AD$ and $EF$ are parallel and divide triangle $ABC$ into three parts of equal area, determine the number of possible integral values for $BC$.

Let $ABC$ be a triangle with $AB < AC$. Its incircle with centre $I$ touches the sides $BC, CA,$ and $AB$ in the points $D, E,$ and $F$ respectively. The angle bisector $AI$ intersects the lines $DE$ and $DF$ in the points $X$ and $Y$ respectively. Let $Z$ be the foot of the altitude through $A$ with respect to $BC$. Prove that $D$ is the incentre of the triangle $XYZ$.

Point $X$ is chosen inside the non trapezoid quadrilateral $ABCD$ such that $\angle AXD +\angle BXC=180$. Suppose the angle bisector of $\angle ABX$ meets the $D$-altitude of triangle $ADX$ in $K$, and the angle bisector of $\angle DCX$ meets the $A$-altitude of triangle $ADX$ in $L$.We know $BK \perp CX$ and $CL \perp BX$. If the circumcenter of $ADX$ is on the line $KL$ prove that $KL \perp AD$. Proposed by [i]Alireza Dadgarnia[/i]

In trapezoid $ABCD$, $AD$ is parallel to $BC$. Knowing that $AB=AD+BC$, prove that the bisector of $\angle A$ also bisects $CD$.

Given a triangle $ABC$, consider the semicircle with diameter $\overline{EF}$ on $\overline{BC}$ tangent to $\overline{AB}$ and $\overline{AC}$. If $BE=1$, $EF=24$, and $FC=3$, find the perimeter of $\triangle{ABC}$. [i]Proposed by Ray Li[/i]

Four lighthouses are located at points $ A$, $ B$, $ C$, and $ D$. The lighthouse at $ A$ is $ 5$ kilometers from the lighthouse at $ B$, the lighthouse at $ B$ is $ 12$ kilometers from the lighthouse at $ C$, and the lighthouse at $ A$ is $ 13$ kilometers from the lighthouse at $ C$. To an observer at $ A$, the angle determined by the lights at $ B$ and $ D$ and the angle determined by the lights at $ C$ and $ D$ are equal. To an observer at $ C$, the angle determined by the lights at $ A$ and $ B$ and the angle determined by the lights at $ D$ and $ B$ are equal. The number of kilometers from $ A$ to $ D$ is given by $ \displaystyle\frac{p\sqrt{r}}{q}$, where $ p$, $ q$, and $ r$ are relatively prime positive integers, and $ r$ is not divisible by the square of any prime. Find $ p\plus{}q\plus{}r$,

Let $ABC$ be a triangle inscribed in a circle and let \[ l_a = \frac{m_a}{M_a} \ , \ \ l_b = \frac{m_b}{M_b} \ , \ \ l_c = \frac{m_c}{M_c} \ , \] where $m_a$,$m_b$, $m_c$ are the lengths of the angle bisectors (internal to the triangle) and $M_a$, $M_b$, $M_c$ are the lengths of the angle bisectors extended until they meet the circle. Prove that \[ \frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3 \] and that equality holds iff $ABC$ is an equilateral triangle.

The sides of triangle $BAC$ are in the ratio $2: 3: 4$. $BD$ is the angle-bisector drawn to the shortest side $AC$, dividing it into segments $AD$ and $CD$. If the length of $AC$ is $10$, then the length of the longer segment of $AC$ is: $\text{(A)} \ 3\frac12 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac57 \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac12$

Characterize all triangles $ABC$ s.t. \[ AI_a : BI_b : CI_c = BC: CA : AB \] where $I_a$ etc. are the corresponding excentres to the vertices $A, B , C$

In triangle $ABC$, the difference between angles $B$ and $C$ is equal to $90^o$, and $AL$ is the angle bisector of triangle $ABC$. The bisector of the exterior angle $A$ of the triangle $ABC$ intersects the line $BC$ at the point $F$. Prove that $AL = AF$. (Alexander Dzyunyak)

Let $ABC$ be a triangle with $\angle ABC = 90^o$ and $AB> BC$. Let $D$ be a point on side $AB$ such that $BD = BC$. Let $E$ be the foot of the perpendicular from $D$ on $AC$, and $F$ the reflection of $B$ wrt $CD$. Show that $EC$ is the bisector of angle $\angle BEF$.

In triangle $ABC$ with $AB>BC$, $BM$ is a median and $BL$ is an angle bisector. The line through $M$ and parallel to $AB$ intersects $BL$ at point $D$, and the line through $L$ and parallel to $BC$ intersects $BM$ at point $E$. Prove that $ED$ is perpendicular to $BL$.

In the acute triangle $ABC$ the side $BC> AB$, and the angle bisector $BL = AB$. On the segment $BL$ there is a point $M$, for which $\angle AML = \angle BCA$. Prove that $AM = LC$.

Let $ABCD$ be a parallelogram such that $AB = 35$ and $BC = 28$. Suppose that $BD \perp BC$. Let $\ell_1$ be the reflection of $AC$ across the angle bisector of $\angle BAD$, and let $\ell_2$ be the line through $B$ perpendicular to $CD$. $\ell_1$ and $\ell_2$ intersect at a point $P$. If $PD$ can be expressed in simplest form as $\frac{m}{n}$, find $m + n$.

Given an acute triangle $ABC$ with $AB <AC$. Points $P$ and $Q$ lie on the angle bisector of $\angle BAC$ so that $BP$ and $CQ$ are perpendicular on that angle bisector. Suppose that point $E, F$ lie respectively at sides $AB$ and $AC$ respectively, in such a way that $AEPF$ is a kite. Prove that the lines $BC, PF$, and $QE$ intersect at one point.

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]

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 $\angle ATI=\angle CTI$.

Consider an isosceles triangle $KL_1L_2$ with $|KL_1|=|KL_2|$ and let $KA, L_1B_1,L_2B_2$ be its angle bisectors. Prove that $\cos \angle B_1AB_2 < \frac35$

In a circle, let $AB$ and $BC$ be chords , with $AB =\sqrt3, BC =3\sqrt3, \angle ABC =60^o$. Find the length of the circle chord that divides angle $ \angle ABC$ in half.

Let $ABC$ be a triangle and let $D, E$ and $F$ be the midpoints of the sides $AB, BC$ and $CA$ respectively. Suppose that the angle bisector of $\angle BDC$ meets $BC$ at the point $M$ and the angle bisector of $\angle ADC$ meets $AC$ at the point $N$. Let $MN$ and $CD$ intersect at $O$ and let the line $EO$ meet $AC$ at $P$ and the line $FO$ meet $BC$ at $Q$. Prove that $CD = PQ$.

$M$ and $N$ are the midpoints of the sides $AB$ and $AC$ of a triangle ABC respectively. $P$ and $Q$ are points on the sides $AB$ and $AC$ respectively such that the bisector of the angle $ACB$ also bisects the angle $MCP$, and the bisector of the angle $ABC$ also bisects the angle $NBQ$. If $AP = AQ$, does it follow that $ABC$ is isosceles? (V Senderov)

Triangle $WSE$ has side lengths $WS=13$, $SE=15$, and $WE=14$. Points $J$ and $H$ lie on $\overline{SE}$ such that $SJ=JH=HE=5$. Let the angle bisector of $\angle{WES}$ intersect $\overline{WH}$ and $\overline{WJ}$ at points $M$ and $T$, respectively. Find the area of quadrilateral $JHMT$.

The side lengths of $\triangle ABC$ are integers with no common factor greater than $1$. Given that $\angle B = 2 \angle C$ and $AB < 600$, compute the sum of all possible values of $AB$. [i]Proposed by Eugene Chen[/i]

In obtuse triangle $ABC$, with the obtuse angle at $A$, let $D$, $E$, $F$ be the feet of the altitudes through $A$, $B$, $C$ respectively. $DE$ is parallel to $CF$, and $DF$ is parallel to the angle bisector of $\angle BAC$. Find the angles of the triangle.

$ABCD$ is a trapezoid with $AB$ parallel to $CD$. The external bisectors of the angles at $ B$ and $C$ intersect at $ P$. The external bisectors of the angles at $ A$ and $D$ intersect at $Q$. Show that the length of $PQ$ is equal to half the perimeter of the trapezoid $ABCD$.