Found problems: 1065
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.
Triangle $ABC$ has a right angle at $B$. Point $D$ lies on side $BC$ such that $3\angle BAD = \angle BAC$. Given $AC=2$ and $CD=1$, compute $BD$.
The nine-point Euler circle of triangle $ABC$ is tangent to the excircles in the angle $A,B,C$ at $Fa, Fb, Fc$ respectively. Prove that $AF_a$ bisects the angle $\angle CAB$ if and only if $AFa$ bisects the angle $\angle F_bAF_c$.
Đỗ Thanh Sơn
Points $K, L$ on sides $AC, CB$ respectively of a triangle $ABC$ are the points of contact of the excircles with the corresponding sides . Prove that the straight line through the midpoints of $KL$ and $AB$
(a) divides the perimeter of triangle $ABC$ in half,
(b) is parallel to the bisector of angle $ACB$.
( L Emelianov)
Let $D$ be a point on side $[AB]$ of triangle $ABC$ with $|AB|=|AC|$ such that $[CD]$ is an angle bisector and $m(\widehat{ABC})=40^\circ$. Let $F$ be a point on the extension of $[AB]$ after $B$ such that $|BC|=|AF|$. Let $E$ be the midpoint of $[CF]$. If $G$ is the intersection of lines $ED$ and $AC$, what is $m(\widehat{FBG})$?
$
\textbf{(A)}\ 150^\circ
\qquad\textbf{(B)}\ 135^\circ
\qquad\textbf{(C)}\ 120^\circ
\qquad\textbf{(D)}\ 105^\circ
\qquad\textbf{(E)}\ \text{None of above}
$
Let $AD$ be the bisector of triangle $ABC$. Circle $\omega$ passes through the vertex $A$ and touches the side $BC$ at point $D$. This circle intersects the sides $AC$ and $AB$ for the second time at points $M$ and $N$ respectively. Lines $BM$ and $CN$ intersect the circle for the second time $\omega$ at points $P$ and $Q$, respectively. Lines $AP$ and $AQ$ intersect side $BC$ at points $K$ and $L$, respectively. Prove that $KL=\frac12 BC$
Let $ABC$ be a triangle such that $AC=2AB$. Let $D$ be the point of intersection of the angle bisector of the angle $CAB$ with $BC$. Let $F$ be the point of intersection of the line parallel to $AB$ passing through $C$ with the perpendicular line to $AD$ passing through $A$. Prove that $FD$ passes through the midpoint of $AC$.
The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P,$ and its radius is 21. Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.
Triangle $ABC$ is inscribed in circle $\omega$. A circle with chord $BC$ intersects segments $AB$ and $AC$ again at $S$ and $R$, respectively. Segments $BR$ and $CS$ meet at $L$, and rays $LR$ and $LS$ intersect $\omega$ at $D$ and $E$, respectively. The internal angle bisector of $\angle BDE$ meets line $ER$ at $K$. Prove that if $BE = BR$, then $\angle ELK = \tfrac{1}{2} \angle BCD$.
[i]Proposed by Evan Chen[/i]
Two not necessarily equal non-intersecting wooden disks, one gray and one black, are glued to a plane. An in finite angle with one gray side and one black side can be moved along the plane so that the disks remain outside the angle, while the colored sides of the angle are tangent to the disks of the same color (the tangency points are not the vertices). Prove that it is possible to draw a ray in the angle, starting from the vertex of the angle and such that no matter how the angle is positioned, the ray passes through some fixed point of the plane.
(Egor Bakaev, Ilya Bogdanov, Pavel Kozhevnikov, Vladimir Rastorguev) (Junior version [url=https://artofproblemsolving.com/community/c6h2094701p15140671]here[/url])
[hide=note]There was a mistake in the text of the problem 3, we publish here the correct version. The solutions were estimated according to the text published originally.[/hide]
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]
In triangle $ABC$, $AL$ bisects angle $A$ and $CM$ bisects angle $C$. Points $L$ and $M$ are on $BC$ and $AB$, respectively. The sides of triangle $ABC$ are $a,b,$ and $c$. Then $\frac{\overline{AM}}{\overline{MB}}=k\frac{\overline{CL}}{\overline{LB}}$ where $k$ is:
$ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{bc}{a^2}\qquad\textbf{(C)}\ \frac{a^2}{bc}\qquad\textbf{(D)}\ \frac{c}{b}\qquad\textbf{(E)}\ \frac{c}{a} $
Let $ABC$ be an equilateral triangle. $M$ is the midpoint of segment $AB$ and $N$ is the midpoint of segment $BC$. Let $P$ be the point outside $ABC$ such that the triangle $ACP$ is isosceles and right in $P$. $PM$ and $AN$ are cut in $I$. Prove that $CI$ is the bisector of the angle $MCA$ .
Given a triangle $ \,ABC,\,$ let $ \,I\,$ be the center of its inscribed circle. The internal bisectors of the angles $ \,A,B,C\,$ meet the opposite sides in $ \,A^{\prime },B^{\prime },C^{\prime }\,$ respectively. Prove that
\[ \frac {1}{4} < \frac {AI\cdot BI\cdot CI}{AA^{\prime }\cdot BB^{\prime }\cdot CC^{\prime }} \leq \frac {8}{27}.
\]
An arbitrary point $M$ is selected in the interior of the segment $AB$. The square $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with segments $AM$ and $MB$ as their respective bases. The circles circumscribed about these squares, with centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$.
a) Prove that $N$ and $N'$ coincide;
b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$;
c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.
Let $ABC$ be a triangle with $\angle ABC=120^o$ and triangle bisectors $(AA_1),(BB_1),(CC_1)$, respectively. $B_1F \perp A_1C_1$, where $F\in (A_1C_1)$. Let $R,I$ and $S$ be the centers of the circles which are inscribed in triangles $C_1B_1F,C_1B_1A_1, A_1B_1F$, and $B_1S\cap A_1C_1=\{Q\}$. Show that $R,I,S,Q$ are on the same circle.
In the quadrilateral $ABCD$ the condition $AD = AB + CD$ is fulfilled. The bisectors of the angles $BAD$ and $ADC$ intersect at the point $P $, as shown in Fig. Prove that $BP = CP$.
[img]https://cdn.artofproblemsolving.com/attachments/3/1/67268635aaef9c6dc3363b00453b327cbc01f3.png[/img]
(Maria Rozhkova)
Let $AC>BC$ in $\triangle ABC$ and $M$ and $N$ be the midpoints of $AC$ and $BC$ respectively. The angle bisector of $\angle B$ intersects $\overline{MN}$ at $P$. The incircle of $\triangle ABC$ has center $I$ and touches $BC$ at $Q$. The perpendiculars from $P$ and $Q$ to $MN$ and $BC$ respectively intersect at $R$. Let $S=AB\cap RN$.
a) Prove that $PCQI$ is cyclic
b) Express the length of the segment $BS$ with $a$, $b$, $c$ - the side lengths of $\triangle ABC$ .
Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$.
(Here we always assume that an angle bisector is a ray.)
[i]Proposed by Sergey Berlov, Russia[/i]
In a triangle $ABC$ , $D$ is a point on $BC$ such that $AD$ is the internal bisector of $\angle A$ . Now Suppose $\angle B$=$2\angle C$ and $CD=AB$ . Prove that $\angle A=72^0$.
Let $ ABC$ be a triangle with circum-circle $ \Gamma$. Let $ M$ be a point in the interior of triangle $ ABC$ which is also on the bisector of $ \angle A$. Let $ AM, BM, CM$ meet $ \Gamma$ in $ A_{1}, B_{1}, C_{1}$ respectively. Suppose $ P$ is the point of intersection of $ A_{1}C_{1}$ with $ AB$; and $ Q$ is the point of intersection of $ A_{1}B_{1}$ with $ AC$. Prove that $ PQ$ is parallel to $ BC$.
In a triangle $ ABC$, the bisector of $ \angle ACB$ cuts the side $ AB$ at $ D$. An arbitrary circle $ (O)$ passing through $ C$ and $ D$ meets the lines $ BC$ and $ AC$ at $ M$ and $ N$ (different from $ C$), respectively.
(a) Prove that there is a circle $ (S)$ touching $ DM$ at $ M$ and $ DN$ at $ N$.
(b) If circle $ (S)$ intersects the lines $ BC$ and $ CA$ again at $ P$ and $ Q$ respectively, prove that the lengths of the segments $ MP$ and $ NQ$ are constant as $ (O)$ varies.
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.
Let $D$, $E$, $F$ be the midpoints of the arcs $BC$, $CA$, $AB$ on the circumcircle of a triangle $ABC$ not containing the points $A$, $B$, $C$, respectively. Let the line $DE$ meets $BC$ and $CA$ at $G$ and $H$, and let $M$ be the midpoint of the segment $GH$. Let the line $FD$ meet $BC$ and $AB$ at $K$ and $J$, and let $N$ be the midpoint of the segment $KJ$.
a) Find the angles of triangle $DMN$;
b) Prove that if $P$ is the point of intersection of the lines $AD$ and $EF$, then the circumcenter of triangle $DMN$ lies on the circumcircle of triangle $PMN$.
Let $ABC$ be an isosceles triangle with apex at $C.$ Let $D$ and $E$ be two points on the sides $AC$ and $BC$ such that the angle bisectors $\angle DEB$ and $\angle ADE$ meet at $F,$ which lies on segment $AB.$ Prove that $F$ is the midpoint of $AB.$