Triangle $ABC$ has $AB = 4$, $AC = 5$, and $BC = 6$. An angle bisector is drawn from angle $A$, and meets $BC$ at $M$. What is the nearest integer to $100 \frac{AM}{CM}$?

In an acute triangle $ABC$, let $D$ be a point in $BC$ such that $AD$ is the angle bisector of $\angle{BAC}$. Let $E \neq B$ be the point of intersection of the circumcircle of triangle $ABD$ with the line perpendicular to $AD$ drawn through $B$. Let $O$ be the circumcenter of triangle $ABC$. Prove that $E$, $O$, and $A$ are collinear.

In a triangle $ABC$, with $AB \ne BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \ne B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$.

Let $ABC$ be a scalene (i.e. non-isosceles) triangle. Let $U$ be the center of the circumcircle of this triangle and $I$ the center of the incircle. Assume that the second point of intersection different from $C$ of the angle bisector of $\gamma = \angle ACB$ with the circumcircle of $ABC$ lies on the perpendicular bisector of $UI$. Show that $\gamma$ is the second-largest angle in the triangle $ABC$.

$AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC+AD-AB-AC = 0$, find the range of $\angle BAC$. [i]Alternative formulation.[/i] Let $AD$ be the altitude of triangle $ABC$ to the side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.

Given a convex quadrilateral $ABCD$.Let $\ell_A,\ell_B,\ell_C,\ell_D$ be exterior angle bisectors of quadrilateral $ABCD$. Let $\ell_A \cap \ell_B=K,\ell_B \cap \ell_C=L,\ell_C \cap \ell_D=M,\ell_D \cap \ell_A=N$.Prove that if circumcircles of triangles $ABK$ and $CDM$ be externally tangent to each other then circumcircles of the triangles $BCL$ and $DAN$ are externally tangent to each other.(L.Emelyanov)

Let $ABC$ be an acute triangle and $AD$ a height. The angle bissector of $\angle DAC$ intersects $DC$ at $E$. Let $F$ be a point on $AE$ such that $BF$ is perpendicular to $AE$. If $\angle BAE=45º$, find $\angle BFC$.

Let $\triangle ABC$ be a right-angled triangle with $\angle A = 90^{\circ}$, and $AB < AC$. Let points $D, E, F$ be located on side $BC$ such that $AD$ is the altitude, $AE$ is the internal angle bisector, and $AF$ is the median. Prove that $3AD + AF > 4AE$.

In triangle $\vartriangle ABC$, we have marked points $A_1$ on side $BC, B_1$ on side $AC$, and $C_1$ on side $AB$ so that $AA_1$ is an altitude, $BB_1$ is a median, and $CC_1$ is an angle bisector. It is known that $\vartriangle A_1B_1C_1$ is equilateral. Prove that $\vartriangle ABC$ is equilateral too. (Note: A median connects a vertex of a triangle with the midpoint of the opposite side. Thus, for median $BB_1$ we know that $B_1$ is the midpoint of side $AC$ in $\vartriangle ABC$.)

Let $ABC$ be a triangle such that $\angle{ABC} = 2\angle{BCA}$ and $\angle{CAB}>90^\circ$. Let $M$ be the midpoint of $BC$. The line perpendicular to $AC$ that passes through $C$ cuts the line $AB$ at point $D$. Show that $\angle{AMB} = \angle{DMC}$.

Prove that for any nonisosceles triangle $l_1^2>\sqrt3 S>l_2^2$, where $l_1, l_2$ are the greatest and the smallest bisectors of the triangle and $S$ is its area.

Let $ I $ be the incenter of triangle $ ABC $. The incircle touches $ BC, CA, AB$ at points $ P, Q, R $. A circle passing through $ B , C $ is tangent to the circle $I$ at point $ X $, a circle passing through $ C , A $ is tangent to the circle $I$ at point $ Y $, and a circle passing through $ A , B $ is tangent to the circle $I$ at point $ Z $, respectively. Prove that three lines $ PX, QY, RZ $ are concurrent.

The altitude $AA'$, the median $BB'$, and the angle bisector $CC'$ of a triangle $ABC$ are concurrent at point $K$. Given that $A'K = B'K$, prove that $C'K = A'K$.

A cyclic quadrilateral $ABCD$ has circumcircle $\Gamma$, and $AB+BC=AD+DC$. Let $E$ be the midpoint of arc $BCD$, and $F (\neq C)$ be the antipode of $A$ [i]wrt[/i] $\Gamma$. Let $I,J,K$ be the incenter of $\triangle ABC$, the $A$-excenter of $\triangle ABC$, the incenter of $\triangle BCD$, respectively. Suppose that a point $P$ satisfies $\triangle BIC \stackrel{+}{\sim} \triangle KPJ$. Prove that $EK$ and $PF$ intersect on $\Gamma.$

In isosceles triangle $ABC$ ($AB = BC$) one draws the angle bisector $CD$. The perpendicular to $CD$ through the center of the circumcircle of $ABC$ intersects $BC$ at $E$. The parallel to $CD$ through $E$ meets $AB$ at $F$. Show that $BE$ = $FD$. [i]M. Sonkin[/i]

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$ with $ \left|BC\right|>\left|BA\right|$, $ D$ is a point inside the triangle such that $ \angle ABD\equal{}\angle DBC$, $ \angle BDC\equal{}150{}^\circ$ and $ \angle DAC\equal{}60{}^\circ$. What is the measure of $ \angle BAD$? $\textbf{(A)}\ 45 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 75 \qquad\textbf{(E)}\ 80$

The external angle bisector of the angle $A$ of an acute-angled triangle $ABC$ meets the circumcircle of $\vartriangle ABC$ at point $T$. The perpendicular from the orthocenter $H$ of $\vartriangle ABC$ to the line $TA$ meets the line $BC$ at point $P$. The line $TP$ meets the circumcircce of $\vartriangle ABC$ at point $D$. Prove that $AB^2+DC^2=AC^2+BD^2$ A. Voidelevich

Let $ABCD$ be a trapezoid such that $|AB|=9$, $|CD|=5$ and $BC\parallel AD$. Let the internal angle bisector of angle $D$ meet the internal angle bisectors of angles $A$ and $C$ at $M$ and $N$, respectively. Let the internal angle bisector of angle $B$ meet the internal angle bisectors of angles $A$ and $C$ at $L$ and $K$, respectively. If $K$ is on $[AD]$ and $\dfrac{|LM|}{|KN|} = \dfrac 37$, what is $\dfrac{|MN|}{|KL|}$? $ \textbf{(A)}\ \dfrac{62}{63} \qquad\textbf{(B)}\ \dfrac{27}{35} \qquad\textbf{(C)}\ \dfrac{2}{3} \qquad\textbf{(D)}\ \dfrac{5}{21} \qquad\textbf{(E)}\ \dfrac{24}{63} $

Points $C_0$ and $B_0$ are the respective midpoints of sides $AB$ and $AC$ of a non-isosceles acute triangle $ABC$, $O$ is its circumscenter and $H$ is the orthocenter. Lines $BH$ and $OC_0$ intersect at $P$, while lines $CH$ and $OB_0$ intersect at $Q$. $OPHQ$ is rhombus. Prove that points $A$, $P$ and $Q$ are collinear. (Author: L. Emelyanov)

Construct triangle $ABC$, given the altitude and the angle bisector both from $A$, if it is known for the sides of the triangle $ABC$ that $2BC = AB + AC$. (Alexey Karlyuchenko)

Given triangle $ABC$, let $M$ be the midpoint of side $AB$ and $N$ be the midpoint of side $AC$. A circle is inscribed inside quadrilateral $NMBC$, tangent to all four sides, and that circle touches $MN$ at point $X.$ The circle inscribed in triangle $AMN$ touches $MN$ at point $Y$, with $Y$ between $X$ and $N$. If $XY=1$ and $BC=12$, find, with proof, the lengths of the sides $AB$ and $AC$.

It is given that for no side of the triangle from the height drawn to it, the bisector and the median it is impossible to make a triangle. Prove that one of the angles of the triangle is greater than $135^o$

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 $\angle BCD$.

Let $ABC$ be a triangle. Prove that there is a line $\ell$ (in the plane of triangle $ABC$) such that the intersection of the interior of triangle $ABC$ and the interior of its reflection $A'B'C'$ in $\ell$ has area more than $\frac23$ the area of triangle $ABC$.