Let $ ABC$ be a triangle, and $ I$ its incenter. Let the incircle of $ ABC$ touch side $ BC$ at $ D$, and let lines $ BI$ and $ CI$ meet the circle with diameter $ AI$ at points $ P$ and $ Q$, respectively. Given $ BI \equal{} 6, CI \equal{} 5, DI \equal{} 3$, determine the value of $ \left( DP / DQ \right)^2$.

Let $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$.

In the convex quadrilateral $ABCD$ we have that $\angle BCD = \angle ADC \ge 90 ^o$. The bisectors of $\angle BAD$ and $\angle ABC$ intersect in $M$. Prove that if $M \in CD$, then $M$ is the middle of $CD$.

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 $ABCD$ be a cyclic quadrilateral, let the angle bisectors at $A$ and $B$ meet at $E$, and let the line through $E$ parallel to side $CD$ intersect $AD$ at $L$ and $BC$ at $M$. Prove that $LA + MB = LM$.

Let $ABC$ be an acute triangle, with median $AM$, height $AH$ and internal angle bisector $AL$. Suppose that $B, H, L, M, C$ are collinear in that order, and $LH<LM$. Prove that $BC>2AL$.

Let $\vartriangle ABC$ be a triangle with $\angle ABC > \angle BCA \ge 30^o$ . The angle bisectors of $\angle ABC$ and $\angle BCA$ intersect $CA$ and $AB$ at $D$ and $E$ respectively, and $BD$ and $CE$ intersect at $P$. Suppose that $P D = P E$ and the incircle of $\vartriangle ABC$ has unit radius. What is the maximum possible length of $BC$?

In triangle $ABC$, $AB = 1$ and $AC = 2$. Suppose there exists a point $P$ in the interior of triangle $ABC$ such that $\angle PBC = 70^{\circ}$, and that there are points $E$ and $D$ on segments $AB$ and $AC$, such that $\angle BPE = \angle EPA = 75^{\circ}$ and $\angle APD = \angle DPC = 60^{\circ}$. Let $BD$ meet $CE$ at $Q,$ and let $AQ$ meet $BC$ at $F.$ If $M$ is the midpoint of $BC$, compute the degree measure of $\angle MPF.$ [i]Authors: Alex Zhu and Ray Li[/i]

Let $H$ be an orthocenter of an acute triangle $ABC$. Prove that midpoints of $AB$ and $CH$ and intersection point of angle bisectors of $\angle CAH$ and $\angle CBH$ lie on the same line.

In acute triangle $ABC$ the bisector of $\measuredangle A$ meets side $BC$ at $D$. The circle with center $B$ and radius $BD$ intersects side $AB$ at $M$; and the circle with center $C$ and radius $CD$ intersects side $AC$ at $N$. Then it is always true that $ \textbf{(A)}\ \measuredangle CND+\measuredangle BMD-\measuredangle DAC =120^{\circ} \qquad\textbf{(B)}\ AMDN\ \text{is a trapezoid} \qquad\textbf{(C)}\ BC\ \text{is parallel to}\ MN \\ \qquad\textbf{(D)}\ AM-AN=\frac{3(DB-DC)}{2} \qquad\textbf{(E)}\ AB-AC=\frac{3(DB-DC)}{2}$

In $ \triangle ABC$ we have $ AB \equal{} 7$, $ AC \equal{} 8$, and $ BC \equal{} 9$. Point $ D$ is on the circumscribed circle of the triangle so that $ \overline{AD}$ bisects $ \angle BAC$. What is the value of $ AD/CD$? $ \textbf{(A)}\ \frac{9}{8}\qquad \textbf{(B)}\ \frac{5}{3}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \frac{17}{7}\qquad \textbf{(E)}\ \frac{5}{2}$

Two circles intersect at points $A$ and $B$. A common tangent touches the first circle at point $C$ and the second at point $D$. Let $\angle CBD > \angle CAD$. Let the line $CB$ intersect the second circle again at point $E$. Prove that $AD$ bisects the angle $\angle CAE$. (P Kozhevnikov)

Let $\vartriangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $|BF| = |BE|$ and such that $ED$ is the internal angle bisector of $\angle BEC$. Prove that $|BD|= |EF|$ if and only if $|AF| = |EC|$.

Let $ABC$ be an isosceles triangle with $AB=AC$. Consider a variable point $P$ on the extension of the segment $BC$ beyound $B$ (in other words, $P$ lies on the line $BC$ such that the point $B$ lies inside the segment $PC$). Let $r_{1}$ be the radius of the incircle of the triangle $APB$, and let $r_{2}$ be the radius of the $P$-excircle of the triangle $APC$. Prove that the sum $r_{1}+r_{2}$ of these two radii remains constant when the point $P$ varies. [i]Remark.[/i] The $P$-excircle of the triangle $APC$ is defined as the circle which touches the side $AC$ and the [i]extensions[/i] of the sides $AP$ and $CP$.

In the triangle $ABC$, $\angle .ACB = 60^o$, and the bisectors $AA_1$ and $BB_1$ intersect at the point $M$. Prove that $MB_1 = MA_1$.

Let $ABC$ be an equilateral triangle whose angle bisectors of $B$ and $C$ intersect at $D$. Perpendicular bisectors of $BD$ and $CD$ intersect $BC$ at points $E$ and $Z$ respectively. a) Prove that $BE=EZ=ZC$. b) Find the ratio of the areas of the triangles $BDE$ to $ABC$

Consider the triangle $ABC$, where $\angle B= 30^o, \angle C = 15^o$, and $M$ is the midpoint of the side $[BC]$. Let point $N \in (BC)$ be such that $[NC] = [AB]$. Show that $[AN$ is the angle bisector of $MAC$

Let $ ABCD$ be a cyclic quadrilateral and $ O$ be the intersection of diagonal $ AC$ and $ BD$. The circumcircles of triangle $ ABO$ and the triangle $ CDO$ intersect at $ K$. Let $ L$ be a point such that the triangle $ BLC$ is similar to $ AKD$ (in that order). Prove that if $ BLCK$ is a convex quadrilateral, then it has an incircle.

Given is the non-equilateral triangle $A_1A_2A_3$. $B_{ij}$ is the symmetric of $A_i$ wrt the inner bisector of $\angle A_j$. Prove that lines $B_{12}B_{21}$, $B_{13}B_{31}$ and $B_{23}B_{32}$ are parallel.

Let $ABC$ be a triangle with $AB = 16$, $AC = 10$, $BC = 18$. Let $D$ be a point on $AB$ such that $4AD = AB$ and let E be the foot of the angle bisector from $B$ onto $AC$. Let $P$ be the intersection of $CD$ and $BE$. Find the area of the quadrilateral $ADPE$.

$ M$ is the midpoint of base $ BC$ in a trapezoid $ ABCD$. A point $ P$ is chosen on the base $ AD$. The line $ PM$ meets the line $ CD$ at a point $ Q$ such that $ C$ lies between $ Q$ and $ D$. The perpendicular to the bases drawn through $ P$ meets the line $ BQ$ at $ K$. Prove that $ \angle QBC \equal{} \angle KDA$. [i]Proposed by S. Berlov[/i]

Let $ABCD$ be a parallelogram. Let $M$ be a point on the side $AB$ and $N$ be a point on the side $BC$ such that the segments $AM$ and $CN$ have equal lengths and are non-zero. The lines $AN$ and $CM$ meet at $Q$. Prove that the line $DQ$ is the bisector of the angle $\measuredangle ADC$. [i]Alternative formulation.[/i] Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$. Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$.

Let $X$ be an arbitrary point on side $BC$ of triangle ABC. Triangle $T$ formed by the bisectors of the angles $ABC$, $ACB$ and $AXC$. Prove that: a) the circumscribed circle of the triangle $T$ passes through the vertex $A$. b) the orthocenter of triangle $T$ lies on line $BC$. (Dmytro Prokopenko)

We are given a circle $k$ and a point $A$ outside of $k$. Next we draw three lines through $A$: one secant intersecting the circle $k$ at points $B$ and $C$, and two tangents touching the circle$k$ at points $D$ and $E$. Let $F$ be the midpoint of $DE$. Show that the line $DE$ bisects the angle $\angle BFC$.

On the circle $K$ there are given three distinct points $A,B,C$. Construct (using only a straightedge and a compass) a fourth point $D$ on $K$ such that a circle can be inscribed in the quadrilateral thus obtained.