Found problems: 1065
Cirlce $\Omega$ is inscribed in triangle $ABC$ with $\angle BAC=40$. Point $D$ is inside the angle $BAC$ and is the intersection of exterior bisectors of angles $B$ and $C$ with the common side $BC$. Tangent form $D$ touches $\Omega$ in $E$. FInd $\angle BEC$.
$\triangle{ABC}$ is isosceles with $AB = AC >BC$. Let $D$ be a point in its interior such that $DA = DB+DC$. Suppose that the perpendicular bisector of $AB$ meets the external angle bisector of $\angle{ADB}$ at $P$, and let $Q$ be the intersection of the perpendicular bisector of $AC$ and the external angle bisector of $\angle{ADC}$. Prove that $B,C,P,Q$ are concyclic.
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 $3AD + AF > 4AE$.
On the median $AD$ of the isosceles triangle $ABC$, point $E$ is marked. Point $F$ is the projection of point $E$ on the line $BC$, point $M$ lies on the segment $EF$, points $N$ and $P$ are projections of point $M$ on the lines $AC$ and $AB$, respectively. Prove that the bisectors of the angles $PMN$ and $PEN$ are parallel.
Let $ABC$ be a triangle inscribed in a circle and let
\[ l_a = \frac{m_a}{M_a} \ , \ \ l_b = \frac{m_b}{M_b} \ , \ \ l_c = \frac{m_c}{M_c} \ , \]
where $m_a$,$m_b$, $m_c$ are the lengths of the angle bisectors (internal to the triangle) and $M_a$, $M_b$, $M_c$ are the lengths of the angle bisectors extended until they meet the circle. Prove that
\[ \frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3 \]
and that equality holds iff $ABC$ is an equilateral triangle.
Triangle $ ABC$ has $ AC \equal{} 450$ and $ BC \equal{} 300$. Points $ K$ and $ L$ are located on $ \overline{AC}$ and $ \overline{AB}$ respectively so that $ AK \equal{} CK$, and $ \overline{CL}$ is the angle bisector of angle $ C$. Let $ P$ be the point of intersection of $ \overline{BK}$ and $ \overline{CL}$, and let $ M$ be the point on line $ BK$ for which $ K$ is the midpoint of $ \overline{PM}$. If $ AM \equal{} 180$, find $ LP$.
$M$ and $N$ are the midpoints of the sides $AB$ and $AC$ of a triangle ABC respectively. $P$ and $Q$ are points on the sides $AB$ and $AC$ respectively such that the bisector of the angle $ACB$ also bisects the angle $MCP$, and the bisector of the angle $ABC$ also bisects the angle $NBQ$. If $AP = AQ$, does it follow that $ABC$ is isosceles?
(V Senderov)
Let $ABC$ be a triangle with $AB>AC$. Let $D$ be the foot of the internal angle bisector of $A$. Points $F$ and $E$ are on $AC,AB$ respectively such that $B,C,F,E$ are concyclic. Prove that the circumcentre of $DEF$ is the incentre of $ABC$ if and only if $BE+CF=BC$.
Triangle $ABC$ has side lengths $AB=18$, $BC=36$, and $CA=24$. The circle $\Gamma$ passes through point $C$ and is tangent to segment $AB$ at point $A$.
Let $X$, distinct from $C$, be the second intersection of $\Gamma$ with $BC$. Moreover, let $Y$ be the point on $\Gamma$ such that segment $AY$ is an angle bisector of $\angle XAC$.
Suppose the length of segment $AY$ can be written in the form $AY=\frac{p\sqrt{r}}{q}$ where $p$, $q$, and $r$ are positive integers such that $gcd(p, q)=1$ and $r$ is square free.
Find the value of $p+q+r$.
Triangle $ ABC$ has incentre $ I$ and the incircle touches $ BC, CA$ at $ D, E$ respectively. Let $ BI$ meet $ DE$ at $ G$. Show that $ AG$ is perpendicular to $ BG$.
$D$ and $E$ are points on the sides $BC$ and $AC$ of a triangle $ABC$ such that $AD$ and $BE$ are angle bisectors of the triangle $ABC$. If $DE$ bisects $\angle ADC$, find $\angle A$.
(SI Tokarev)
The triangle $ABC$ has a right angle at $A$. The centre of the circumcircle is called $O$, and the base point of the normal from $O$ to $AC$ is called $D$. The point $E$ lies on $AO$ with $AE = AD$. The angle bisector of $\angle CAO$ meets $CE$ in $Q$. The lines $BE$ and $OQ$ intersect in $F$. Show that the lines $CF$ and $OE$ are parallel.
Let $k$ be a circumcircle of triangle $ABC$ $(AC<BC)$. Also, let $CL$ be an angle bisector of angle $ACB$ $(L \in AB)$, $M$ be a midpoint of arc $AB$ of circle $k$ containing the point $C$, and let $I$ be an incenter of a triangle $ABC$. Circle $k$ cuts line $MI$ at point $K$ and circle with diameter $CI$ at $H$. If the circumcircle of triangle $CLK$ intersects $AB$ again at $T$, prove that $T$, $H$ and $C$ are collinear.
.
In an acute triangle $ABC$ , the bisector of angle $\angle A$ intersects the circumscribed circle of the triangle $ABC$ at the point $W$. From point $W$ , a parallel is drawn to the side $AB$, which intersects this circle at the point $F \ne W$. Describe the construction of the triangle $ABC$, if given are the segments $FA$ , $FW$ and $\angle FAC$.
(Andrey Mostovy)
Let $M$ and $N$ be touching points of incircle with sides $AB$ and $AC$ of triangle $ABC$, and $P$ intersection point of line $MN$ and angle bisector of $\angle ABC$. Prove that $\angle BPC =90 ^{\circ}$
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$.
Show that in a right-angled triangle the bisector of the right angle divides into equal parts the angle between the altitude and the median, drawn from the same vertex.
In the acute-angled triangle $ABC$, the segment $AP$ was drawn and the center was marked $O$ of the circumscribed circle. The circumcircle of triangle $ABP$ intersects the line $AC$ for the second time at point $X$, the circumcircle of the triangle $ACP$ intersects the line $AB$ for the second time at the point $Y$. Prove that the lines $XY$ and $PO$ are perpendicular if and only if $P$ is the foor of the bisector of the triangle $ABC$.
$\triangle{ABC}$ is isosceles$(AB=AC)$. Points $P$ and $Q$ exist inside the triangle such that $Q$ lies inside $\widehat{PAC}$ and $\widehat{PAQ} = \frac{\widehat{BAC}}{2}$. We also have $BP=PQ=CQ$.Let $X$ and $Y$ be the intersection points of $(AP,BQ)$ and $(AQ,CP)$ respectively. Prove that quadrilateral $PQYX$ is cyclic. [i](20 Points)[/i]
$(NET 5)$ The bisectors of the exterior angles of a pentagon $B_1B_2B_3B_4B_5$ form another pentagon $A_1A_2A_3A_4A_5.$ Construct $B_1B_2B_3B_4B_5$ from the given pentagon $A_1A_2A_3A_4A_5.$
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 $ABC$.
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 $ ABCD$ be an isosceles trapezoid with $ AD \parallel BC$. Its diagonals $ AC$ and $ BD$ intersect at point $ M$. Points $ X$ and $ Y$ on the segment $ AB$ are such that $ AX \equal{} AM$, $ BY \equal{} BM$. Let $ Z$ be the midpoint of $ XY$ and $ N$ is the point of intersection of the segments $ XD$ and $ YC$. Prove that the line $ ZN$ is parallel to the bases of the trapezoid.
[i]Author: A. Akopyan, A. Myakishev[/i]
The orthocenter of an acute triangle $ABC$ is $H$. Let $K$ and $P$ be the midpoints of lines $BC$ and $AH$, respectively. The angle bisector drawn from the vertex $A$ of the triangle $ABC$ intersects with line $KP$ at $D$. Prove that $HD\perp AD$.
Let $AH$, $BL$ and $CM$ be an altitude, a bisectrix and a median in triangle $ABC$. It is known that lines $AH$ and $BL$ are an altitude and a bisectrix of triangle $HLM$. Prove that line $CM$ is a median of this triangle.