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]

In convex quadrilateral $ ABCD$, $ AB \equal{} BC$ and $ AD \equal{} DC$. Point $ E$ lies on segment $ AB$ and point $ F$ lies on segment $ AD$ such that $ B$, $ E$, $ F$, $ D$ lie on a circle. Point $ P$ is such that triangles $ DPE$ and $ ADC$ are similar and the corresponding vertices are in the same orientation (clockwise or counterclockwise). Point $ Q$ is such that triangles $ BQF$ and $ ABC$ are similar and the corresponding vertices are in the same orientation. Prove that points $ A$, $ P$, $ Q$ are collinear.

In triangle $ABC$, the point $H$ is the orthocenter. A circle centered at point $H$ and with radius $AH$ intersects the lines $AB$ and $AC$ at points $E$ and $D$, respectively. The point $X$ is the symmetric of the point $A$ with respect to the line $BC$ . Prove that $XH$ is the bisector of the angle $DXE$. (Matthew of Kursk)

We consider the quadrilateral $ABCD$, with $AD = CD = CB$ and $AB \parallel CD$. Points $E$ and $F$ belong to the segments $CD$ and $CB$ so that angles $\angle ADE = \angle AEF$. Prove that: a) $4CF \le CB$ , b) if $4CF = CB$, then $AE$ is the bisector of the angle $\angle DAF$.

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)

In quadrilateral $ABCD$ sides $BC$ and $AD$ are parallel. In each of the four vertices we draw an angular bisector. The angular bisectors of angles $A$ and $B$ intersect in point $P$, those of angles $B$ and $C$ intersect in point $Q$, those of angles $C$ and $D$ intersect in point $R$, and those of angles $D$ and $A$ intersect in point S. Suppose that $PS$ is parallel to $QR$. Prove that $|AB| =|CD|$. [asy] unitsize(1.2 cm); pair A, B, C, D, P, Q, R, S; A = (0,0); D = (3,0); B = (0.8,1.5); C = (3.2,1.5); S = extension(A, incenter(A,B,D), D, incenter(A,C,D)); Q = extension(B, incenter(A,B,C), C, C + incenter(A,B,D) - A); P = extension(A, S, B, Q); R = extension(D, S, C, Q); draw(A--D--C--B--cycle); draw(B--Q--C); draw(A--S--D); dot("$A$", A, SW); dot("$B$", B, NW); dot("$C$", C, NE); dot("$D$", D, SE); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(270)); dot("$R$", R, dir(90)); dot("$S$", S, dir(90)); [/asy] Attention: the figure is not drawn to scale.

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.

A parallelogram $ABCD$ is given ($AB \neq BC$). Points $E$ and $G$ are chosen on the line $\overline{CD}$ such that $\overline{AC}$ is the angle bisector of both angles $\angle EAD$ and $\angle BAG$. The line $\overline{BC}$ intersects $\overline{AE}$ and $\overline{AG}$ at $F$ and $H$, respectively. Prove that the line $\overline{FG}$ passes through the midpoint of $HE$. [i]Proposed by Mahdi Etesamifard[/i]

In a triangle $ABC$ the point $D$ is the intersection of the interior angle bisector of $\angle BAC$ and side $BC$. Let $P$ be the second intersection point of the exterior angle bisector of $\angle BAC$ with the circumcircle of $\angle ABC$. A circle through $A$ and $P$ intersects line segment $BP$ internally in $E$ and line segment $CP$ internally in $F$. Prove that $\angle DEP = \angle DFP$.

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 $OA_1=OB_1$

Median $AM$ and the angle bisector $CD$ of a right triangle $ABC$ ($\angle B=90^o$) intersect at the point $O$. Find the area of the triangle $ABC$ if $CO=9, OD=5$.

As shown in the figure below, the in-circle of $ABC$ is tangent to sides $AB$ and $AC$ at $D$ and $E$ respectively, and $O$ is the circumcenter of $BCI$. Prove that $\angle ODB = \angle OEC$. [asy]import graph; size(5.55cm); pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-5.76,xmax=4.8,ymin=-3.69,ymax=3.71; pen zzttqq=rgb(0.6,0.2,0), wwwwqq=rgb(0.4,0.4,0), qqwuqq=rgb(0,0.39,0); pair A=(-2,2.5), B=(-3,-1.5), C=(2,-1.5), I=(-1.27,-0.15), D=(-2.58,0.18), O=(-0.5,-2.92); D(A--B--C--cycle,zzttqq); D(arc(D,0.25,-104.04,-56.12)--(-2.58,0.18)--cycle,qqwuqq); D(arc((-0.31,0.81),0.25,-92.92,-45)--(-0.31,0.81)--cycle,qqwuqq); D(A--B,zzttqq); D(B--C,zzttqq); D(C--A,zzttqq); D(CR(I,1.35),linewidth(1.2)+dotted+wwwwqq); D(CR(O,2.87),linetype("2 2")+blue); D(D--O); D((-0.31,0.81)--O); D(A); D(B); D(C); D(I); D(D); D((-0.31,0.81)); D(O); MP( "A", A, dir(110)); MP("B", B, dir(140)); D("C", C, dir(20)); D("D", D, dir(150)); D("E", (-0.31, 0.81), dir(60)); D("O", O, dir(290)); D("I", I, dir(100)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

At the base of the triangular pyramid $ABCD$ lies a regular triangle $ABC$ such that $AD = BC$. All plane angles at vertex $B$ are equal to each other. What might these angles be equal to?

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.

Find all triangles $ABC$ for which $\angle ACB$ is acute and the interior angle bisector of $BC$ intersects the trisectors $(AX, (AY$ of the angle $\angle BAC$ in the points $N,P$ respectively, such that $AB=NP=2DM$, where $D$ is the foot of the altitude from $A$ on $BC$ and $M$ is the midpoint of the side $BC$.

Points $D$ and $E$ are taken on $[BC]$ and $[AC]$ of acute angled triangle $ABC$ such that $BD$ and $CE$ are angle bisectors. Projections of $D$ onto $BC$ and $BA$ are $P$ and $Q$, projections of $E$ onto $CA$ and $CB$ are $R$ and $S$. Let $AP \cap CQ=X$, $AS \cap BR=Y$ and $BX \cap CY=Z$. Show that $AZ \perp BC$.

In triangle $ABC$, bisectors are drawn $AA_1$ and $CC_1$. Prove that if the length of the perpendiculars drawn from the vertex $B$ on lines $AA1$ and $CC_1$ are equal, then $\vartriangle ABC$ is isosceles.

9.4, 10.3 Let $I$ be an incenter of $ABC$ ($AB<BC$), $M$ is a midpoint of $AC$, $N$ is a midpoint of circumcircle's arc $ABC$. Prove that $\angle IMA=\angle INB$. ([i]A. Badzyan[/i])

Let $ABC$ be a triangle with $|AB|> |BC|$. Let $D$ be the midpoint of $AC$. Let $E$ be the intersection of the angular bisector of $\angle ABC$ and the line $AC$. Let $F$ be the point on $BE$ such that $CF$ is perpendicular to $BE$. Finally, let $G$ be the intersection of $CF$ and $BD$. Prove that $DF$ divides the line segment $EG$ into two equal parts.

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.

Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?

Consider a triangle $\triangle ABC$, let the incircle centered at $I$ touch the sides $BC,CA,AB$ at points $D,E,F$ respectively. Let the angle bisector of $\angle BIC$ meet $BC$ at $M$, and the angle bisector of $\angle EDF$ meet $EF$ at $N$. Prove that $A,M,N$ are collinear.

Let $ABC$ be an acute-angled triangle with $AC > BC$, with incircle $\tau$ centered at $I$ which touches $BC$ and $AC$ at points $D$ and $E$, respectively. The point $M$ on $\tau$ is such that $BM \parallel DE$ and $M$ and $B$ lie on the same halfplane with respect to the angle bisector of $\angle ACB$. Let $F$ and $H$ be the intersections of $\tau$ with $BM$ and $CM$ different from $M$, respectively. Let $J$ be a point on the line $AC$ such that $JM \parallel EH$. Let $K$ be the intersection of $JF$ and $\tau$ different from $F$. Prove that $ME \parallel KH$.

Let $ABCD$ be an isosceles trapezoid with bases $AD> BC$. Let $X$ be the intersection of the bisectors of $\angle BAC$ and $BC$. Let $E$ be the intersection of$ DB$ with the parallel to the bisector of $\angle CBD$ through $X$ and let $F$ be the intersection of $DC$ with the parallel to the bisector of $\angle DCB$ through $X$. Show that quadrilateral $AEFD$ is cyclic.

Given a parabola $C: y^2=4px\ (p>0)$ with focus $F(p,\ 0)$. Let two lines $l_1,\ l_2$ passing through $F$ intersect orthogonaly each other, $C$ intersects with $l_1$ at two points $P_1,\ P_2$ and $C$ intersects with $l_2$ at two points $Q_1,\ Q_2$. Answer the following questions. (1) Set the equation of $l_1$ as $x=ay+p$ and let the coordinates of $P_1,\ P_2$ as $(x_1,\ y_1),\ (x_2,\ y_2)$, respectively. Express $y_1+y_2,\ y_1y_2$ in terms of $a,\ p$. (2) Show that $\frac{1}{P_1P_2}+\frac{1}{Q_1Q_2}$ is constant regardless of way of taking $l_1,\ l_2$.