Found problems: 1065
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)
In an acute-angled triangle $ ABC,$ let $ AD,BE$ be altitudes and $ AP,BQ$ internal bisectors. Denote by $ I$ and $ O$ the incenter and the circumcentre of the triangle, respectively. Prove that the points $ D, E,$ and $ I$ are collinear if and only if the points $ P, Q,$ and $ O$ are collinear.
Let triangle $ABC$ be an isosceles triangle with $AB = AC$. Suppose that the angle bisector of its angle $\angle B$ meets the side $AC$ at a point $D$ and that $BC = BD+AD$.
Determine $\angle A$.
Given an acute triangle $ABC$ with their incenter $I$. Point $X$ lies on $BC$ on the same side as $B$ wrt $AI$. Point $Y$ lies on the shorter arc $AB$ of the circumcircle $ABC$. It is given that $$\angle AIX = \angle XYA = 120^\circ.$$
Prove that $YI$ is the angle bisector of $XYA$.
Let $I$ be the incenter of the scalene $\Delta ABC$, such, $AB<AC$, and let $I'$ be the reflection of point $I$ in line $BC$. The angle bisector $AI$ meets $BC$ at $D$ and circumcircle of $\Delta ABC$ at $E$. The line $EI'$ meets the circumcircle at $F$. Prove, that,
$\text{(i) } \frac{AI}{IE}=\frac{ID}{DE}$
$\text{(ii) } IA=IF$
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}$
Let $OX$ and $OY$ be non-collinear rays. Through a point $A$ on $OX$, draw two lines $r_1$ and $r_2$ that are antiparallel with respect to $\angle XOY$. Let $r_1$ cut $OY$ at $M$ and $r_2$ cut $OY$ at $N$. (Thus, $\angle OAM = \angle ONA$). The bisectors of $ \angle AMY$ and $\angle ANY$ meet at $P$. Determine the location of $P$.
Let $ABC$ be a triangle, $P$ and $Q$ the intersections of the parallel line to $BC$ that passes through $A$ with the external angle bisectors of angles $B$ and $C$, respectively. The perpendicular to $BP$ at $P$ and the perpendicular to $CQ$ at $Q$ meet at $R$. Let $I$ be the incenter of $ABC$. Show that $AI = AR$.
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
$\triangle{ABC}$ 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 $\Omega$ 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]
Let $ ABC $ be a triangle and the point $ D $ is on the segment $ BC $ such that $ AD $ is the interior bisector of $ \angle A $. We stretch $ AD $ such that it meets the circumcircle of $ \Delta ABC $ at $ M $. We draw a line from $ D $ such that it meets the lines $ MB,MC $ at $ P,Q $, respectively ($ M $ is not between $ B,P $ and also is not between $ C,Q $).
Prove that $ \angle PAQ\geq\angle BAC $.
Let $ABCD$ be a convex quadrilateral with $A,B,C,D$ concyclic. Assume $\angle ADC$ is acute and $\frac{AB}{BC}=\frac{DA}{CD}$. Let $\Gamma$ be a circle through $A$ and $D$, tangent to $AB$, and let $E$ be a point on $\Gamma$ and inside $ABCD$.
Prove that $AE\perp EC$ if and only if $\frac{AE}{AB}-\frac{ED}{AD}=1$.
The angle bisector $[AD]$, the median $[BM]$ and the height $[CH]$ of the acute-angled triangle $ABC$ intersect in one point. Prove that the $\angle BAC> 45^o$.
Let $AM$ and $AL$ be the median and bisector of a triangle $ABC$ ($M,L \in BC$).
If $BC = a, AM = m_a, AL = l_a$, prove the inequalities:
(a) $a\tan \frac{a}{2} \le 2m_a \le a \cot \frac{a}{2} $ if $a < \frac{\pi}{2}$ and $a\tan \frac{a}{2} \ge 2m_a \ge a \cot \frac{a}{2} $ if $a > \frac{\pi}{2}$
(b) $2l_a \le a\cot \frac{a}{2} $.
In the triangle $ABC$, angle bisector $CD$ intersects the circumcircle of $ABC$ at the point $K$.
(a) Prove the equalities:
$$\frac1{ID}-\frac1{IK}=\frac1{CI},\enspace\frac{CI}{ID}-\frac{ID}{DK}=1$$where $I$ is the center of the inscribed circle of triangle $ABC$.
(b) On the segment $CK$ some point $P$ is chosen whose projections on $AC,BC,AB$ respectively are $P_1,P_2,P_3$. The lines $PP_3$ and $P_1P_2$ intersect at a point $M$. Find the locus of $M$ when $P$ moves around segment $CK$.
Let $ABCD$ be a cyclic quadrilateral wih both diagonals perpendicular to each other and intersecting at point $O$. Let $E,F,G,H$ be the orthogonal projections of $O$ on sides $AB,BC,CD,DA$ respectively.
a. Prove that $\angle EFG + \angle GHE = 180^o$
b. Prove that $OE$ bisects angle $\angle FEH$ .
Let $ABC$ be a triangle and let $D, E$ and $F$ be the midpoints of the sides $AB, BC$ and $CA$ respectively. Suppose that the angle bisector of $\angle BDC$ meets $BC$ at the point $M$ and the angle bisector of $\angle ADC$ meets $AC$ at the point $N$. Let $MN$ and $CD$ intersect at $O$ and let the line $EO$ meet $AC$ at $P$ and the line $FO$ meet $BC$ at $Q$. Prove that $CD = PQ$.
Let the $ABCD$ be a quadrilateral without parallel sides, inscribed in a circle. Let $P$ and $Q$ be the intersection points between the lines containing the quadrilateral opposite sides. Show that the bisectors to the angles at $P$ and $Q$ are parallel to the bisectors of the angles at the intersection point of the diagonals of the quadrilateral.
Let $ABCDEF$ be a convex hexagon such that $\angle A = \angle C = \angle E$ and $\angle B = \angle D = \angle F$ and the (interior) angle bisectors of $\angle A, ~\angle C,$ and $\angle E$ are concurrent.
Prove that the (interior) angle bisectors of $\angle B, ~\angle D, $ and $\angle F$ must also be concurrent.
[i]Note that $\angle A = \angle FAB$. The other interior angles of the hexagon are similarly described.[/i]
Let triangle $ABC$ have $\angle BAC = 45^{\circ}$ and circumcircle $\Gamma$ and let $M$ be the intersection of the angle bisector of $\angle BAC$ with $\Gamma$. Let $\Omega$ be the circle tangent to segments $\overline{AB}$ and $\overline{AC}$ and internally tangent to $\Gamma$ at point $T$. Given that $\angle TMA = 45^{\circ}$ and that $TM = \sqrt{100 - 50 \sqrt{2}}$, the length of $BC$ can be written as $a \sqrt{b}$, where $b$ is not divisible by the square of any prime. Find $a + b$.
Let $ABC$ be an acute-angled triangle with circumference $\Omega$. Let the angle bisectors of $\angle B$ and $\angle C$ intersect $\Omega$ again at $M$ and $N$. Let $I$ be the intersection point of these angle bisectors. Let $M'$ and $N'$ be the respective reflections of $M$ and $N$ in $AC$ and $AB$. Prove that the center of the circle passing through $I$, $M'$, $N'$ lies on the altitude of triangle $ABC$ from $A$.
[i]Proposed by Victor Domínguez and Ariel García[/i]
Inside the angle $AOD$, the rays $OB$ and $OC$ are drawn such that $\angle AOB = \angle COD.$ Two circles are inscribed inside the angles $\angle AOB$ and $\angle COD$ . Prove that the intersection point of the common internal tangents of these circles lies on the bisector of the angle $AOD$.
Let $ABC$ be a triangle. The internal bisector of $\angle B$ meets $AC$ in $P$ and $I$ is the incenter of $ABC$. Prove that if $AP+AB = CB$, then $API$ is an isosceles triangle.
Point $M$ is a midpoint of side $BC$ of a triangle $ABC$ and $H$ is the orthocenter of $ABC$. $MH$ intersects the $A$-angle bisector at $Q$. Points $X$ and $Y$ are the projections of $Q$ on sides $AB$ and $AC$. Prove that $XY$ passes through $H$.