Let $\triangle ABC$ be a triangle with $AB<AC$. Let the angle bisector of $\angle BAC$ meet $BC$ at $D$, and let $M$ be the midpoint of $\overline{BC}$. Let $P$ be the foot of the perpendicular from $B$ to $\overline{AD}$. Extend $\overline{BP}$ to meet $\overline{AM}$ at $Q$. Show that $\overline{DQ}$ is parallel to $\overline{AB}$.

In the triangle $ABC, \angle BAC$ is acute, the angle bisector of $\angle BAC$ meets $BC$ at $D, K$ is the foot of the perpendicular from $B$ to $AC$, and $\angle ADB = 45^o$. Point $P$ lies between $K$ and $C$ such that $\angle KDP = 30^o$. Point $Q$ lies on the ray $DP$ such that $DQ = DK$. The perpendicular at $P$ to $AC$ meets $KD$ at $L$. Prove that $PL^2 = DQ \cdot PQ$.

Let $O$ be the circumcenter of the acute triangle $ABC$ ($AB < AC$). Let $A_1$ and $P$ be the feet of the perpendicular lines drawn from $A$ and $O$ to $BC$, respectively. The lines $BO$ and $CO$ intersect $AA_1$ in $D$ and $E$, respectively. Let $F$ be the second intersection point of $\odot ABD$ and $\odot ACE$. Prove that the angle bisector od $\angle FAP$ passes through the incenter of $\triangle ABC$.

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 $C_{1}KM$.

The points $M$ and $N$ are chosen on the angle bisector $AL$ of a triangle $ABC$ such that $\angle ABM=\angle ACN=23^{\circ}$. $X$ is a point inside the triangle such that $BX=CX$ and $\angle BXC=2\angle BML$. Find $\angle MXN$.

In $\Delta ABC$, $AC=kAB$, with $k>1$. The internal angle bisector of $\angle BAC$ meets $BC$ at $D$. The circle with $AC$ as diameter cuts the extension of $AD$ at $E$. Express $\dfrac{AD}{AE}$ in terms of $k$.

The internal bisector of angle $A$ in a triangle $ABC$ with $AC > AB$ meets the circumcircle $\Gamma$ of the triangle in $D$. Join$D$ to the center $O$ of the circle $\Gamma$ and suppose that $DO$ meets $AC$ in $E$, possibly when extended. Given that $BE$ is perpendicular to $AD$, show that $AO$ is parallel to $BD$.

It is given isosceles triangle $ABC$ ($AB=AC$) such that $\angle BAC=108^{\circ}$. Angle bisector of angle $\angle ABC$ intersects side $AC$ in point $D$, and point $E$ is on side $BC$ such that $BE=AE$. If $AE=m$, find $ED$

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$.

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]

Point $M$ is the midpoint of base $AD$ of an isosceles trapezoid $ABCD$ with circumcircle $\omega$. The angle bisector of $ABD$ intersects $\omega$ at $K$. Line $CM$ meets $\omega$ again at $N$. From point $B$, tangents $BP, BQ$ are drawn to $(KMN)$. Prove that $BK, MN, PQ$ are concurrent. [i]A. Kuznetsov[/i]

Let $ABC$ be a triangle with circumscribed and inscribed circles $(O)$ and $(I)$ respectively. $AA'$,$BB'$,$CC'$ are the bisectors of triangle $ABC$. Prove that $OI$ passes through the the isogonal conjugate of point $I$ with respect to triangle $A'B'C'$.

The sides of triangle $ABC$ are $a = | BC |, b = | CA |$ and $c = | AB |$. Points $D, E$ and $F$ are the points on the sides $BC, CA$ and $AB$ such that $AD, BE$ and $CF$ are the angle bisectors of the triangle $ABC$. Determine the lengths of the segments $AD, BE$, and $CF$ in terms of $a, b$, and $c$.

In convex quadrilateral $ ABCD$, $ CB,DA$ are external angle bisectors of $ \angle DCA,\angle CDB$, respectively. Points $ E,F$ lie on the rays $ AC,BD$ respectively such that $ CEFD$ is cyclic quadrilateral. Point $ P$ lie in the plane of quadrilateral $ ABCD$ such that $ DA,CB$ are external angle bisectors of $ \angle PDE,\angle PCF$ respectively. $ AD$ intersects $ BC$ at $ Q.$ Prove that $ P$ lies on $ AB$ if and only if $ Q$ lies on segment $ EF$.

On the extension of the angle bisector $AL$ of the triangle $ABC$, a point $P$ is placed such that $P L = AL$. Prove that the perimeter of triangle $PBC$ does not exceed the perimeter of triangle $ABC$.

In a rectangle $ABCD, M$ and $N$ are the midpoints of $AD$ and $BC$ respectively and $P$ is a point on line $CD$. The line $PM$ meets $AC$ at $Q$. Prove that MN bisects the angle $\angle QNP$.

In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other.

$ABCD$ is a square of side 1. $P$ and $Q$ are points on $AB$ and $BC$ such that $\widehat{PDQ} = 45^{\circ}$. Find the perimeter of $\Delta PBQ$.

In a triangle $ABC$, $\angle B= 2\angle A \ne 90^o$ . The inner bisector of $B$ intersects the perpendicular bisector of $[AC]$ at a point $D$. Prove that $AB \parallel CD$.

In $\vartriangle ABC$, the bisector of $\angle B$ meets $AC$ at $D$ and the bisector of $\angle C$ meets $AB$ at $E$. These bisectors intersect at $O$ and $OD = OE$. If $AD \ne AE$, prove that $\angle A = 60^o$.

Point $A$ lies inside an angle with vertex $M$. A ray issuing from point $A$ is reflected in one side of the angle at point $B$, then in the other side at point $C$ and then returns back to point $A$ (the ordinary rule of reflection holds). Prove that the center of the circle circumscribed about triangle $\triangle BCM$ lies on line $AM$.

Let $ ABCD $ be a convex quadrilateral with no pair of parallel sides, such that $ \angle ABC = \angle CDA $. Assume that the intersections of the pairs of neighbouring angle bisectors of $ ABCD $ form a convex quadrilateral $ EFGH $. Let $ K $ be the intersection of the diagonals of $ EFGH$. Prove that the lines $ AB $ and $ CD $ intersect on the circumcircle of the triangle $ BKD $.

The angle bisectors $AA_1$ and $BB_1$ of the triangle ABC intersect at point $O$. Prove that when the angle $C$ is equal to $60^0$, then $OA_1=OB_1$

Given a convex pentagon $ABCDE$ such that $AE$ is parallel to $CD$ and $AB=BC$. Angle bisectors of angles $A$ and $C$ intersect at $K$. Prove that $BK$ and $AE$ are parallel.

In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$. Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10.$ Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG.$ [asy] size(250); pair A=(0,12), E=(0,8), B=origin, C=(24*sqrt(2),0), D=(6*sqrt(2),0), F=A+10*dir(A--C), G=intersectionpoint(E--F, A--D); draw(A--B--C--A--D^^E--F); pair point=G+1*dir(250); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G)); markscalefactor=0.1; draw(rightanglemark(A,B,C)); label("10", A--F, dir(90)*dir(A--F)); label("27", F--C, dir(90)*dir(F--C)); label("3", (0,10), W); label("9", (0,4), W);[/asy]