Found problems: 1065
Prove that any triangle having two equal internal angle bisectors (each measured from a vertex to the opposite side) is isosceles.
Given a triangle $ABC$ where $AC > BC$, $D$ is located on the circumcircle of $ABC$ such that $D$ is the midpoint of the arc $AB$ that contains $C$. $E$ is a point on $AC$ such that $DE$ is perpendicular to $AC$. Prove that $AE = EC + CB$.
Let $\triangle ABC$ have $\angle A=20^{\circ}$ and $\angle C=40^{\circ}$. We've constructed the angle bisector $AL$ ($L\in BC$) and the external angle bisector $CN$ ($N\in AB$). Find $\angle CLN$.
In a non-isosceles acute triangle $ABC$, $D$ is the midpoint of the edge $[BC]$. The points $E$ and $F$ lie on $[AC]$ and $[AB]$, respectively, and the circumcircles of $CDE$ and $AEF$ intersect in $P$ on $[AD]$. The angle bisector from $P$ in triangle $EFP$ intersects $EF$ in $Q$. Prove that the tangent line to the circumcirle of $AQP$ at $A$ is perpendicular to $BC$.
A triangle with angles of $30, 70$ and $80$ degrees is given. Cut it by a straight line into two triangles in such a way that an angle bisector in one of these triangles and a median in the other one drawn from two endpoints of the cutting segment are parallel to each other. (It suffices to find one such cutting.)
(A. Shapovalov )
The bisectors of the angles $A$ and $B$ of the triangle $ABC$ intersect the sides $BC$ and $AC$ at points $D$ and $E$. It is known that $AE + BD = AB$. Find the angle $\angle C$.
Let a line $\ell$ intersect the line $AB$ at $F$, the sides $AC$ and $BC$ of a triangle $ABC$ at $D$ and $E$, respectively and the internal bisector of the angle $BAC$ at $P$. Suppose that $F$ is at the opposite side of $A$ with respect to the line $BC$, $CD = CE$ and $P$ is in the interior the triangle $ABC$. Prove that
\[FB \cdot FA+CP^2 = CF^2 \iff AD \cdot BE = PD^2.\]
It is given that for no side of the triangle from the height drawn to it, the bisector and the median it is impossible to make a triangle. Prove that one of the angles of the triangle is greater than $135^o$
The diagonals $ AC$ and $ BD$ of a convex quadrilateral $ ABCD$ intersect at point $ M$. The bisector of $ \angle ACD$ meets the ray $ BA$ at $ K$. Given that $ MA \cdot MC \plus{}MA \cdot CD \equal{} MB \cdot MD$, prove that $ \angle BKC \equal{} \angle CDB$.
The points $A, B, C, D$ lie in this order on a circle $o$. The point $S$ lies inside $o$ and has properties $\angle SAD=\angle SCB$ and $\angle SDA= \angle SBC$. Line which in which angle bisector of $\angle ASB$ in included cut the circle in points $P$ and $Q$. Prove $PS =QS$.
Consider $\omega$ is circumcircle of an acute triangle $ABC$. $D$ is midpoint of arc $BAC$ and $I$ is incenter of triangle $ABC$. Let $DI$ intersect $BC$ in $E$ and $\omega$ for second time in $F$. Let $P$ be a point on line $AF$ such that $PE$ is parallel to $AI$. Prove that $PE$ is bisector of angle $BPC$.
[i]Proposed by Mr.Etesami[/i]
Angle bisector at $A$, altitude from $B$ to $CA$ and altitude of $C$ to $AB$ on a scalene triangle $ABC$ forms a triangle $\triangle$. Let $P$ and $Q$ points on lines $AB$ and $AC$, respectively, such that the midpoint of segment $PQ$ is the orthocenter of the triangle $\triangle$. Prove that the points $B, C, P$ and $Q$ lie on a circle.
A median , a bisector and an altitude of a certain triangle intersect at an inner point $O$ . The segment of the bisector from the vertex to $O$ is equal to the segment of the altitude from the vertex to $O$ . Prove that the triangle is equilateral .
Prove the following two statements:
(a) If a triangle is isosceles, then two of its bisectors are of equal length.
(b) If two angle bisectors in a triangle are of equal length, then it is isosceles.
Let $ ABC$ be a triangle; $ \Gamma_A,\Gamma_B,\Gamma_C$ be three equal, disjoint circles inside $ ABC$ such that $ \Gamma_A$ touches $ AB$ and $ AC$; $ \Gamma_B$ touches $ AB$ and $ BC$; and $ \Gamma_C$ touches $ BC$ and $ CA$. Let $ \Gamma$ be a circle touching circles $ \Gamma_A, \Gamma_B, \Gamma_C$ externally. Prove that the line joining the circum-centre $ O$ and the in-centre $ I$ of triangle $ ABC$ passes through the centre of $ \Gamma$.
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?
Let $ABC$ be a scalene (i.e. non-isosceles) triangle. Let $U$ be the center of the circumcircle of this triangle and $I$ the center of the incircle. Assume that the second point of intersection different from $C$ of the angle bisector of $\gamma = \angle ACB$ with the circumcircle of $ABC$ lies on the perpendicular bisector of $UI$.
Show that $\gamma$ is the second-largest angle in the triangle $ABC$.
Three angle bisectors were drawn in a triangle, and it turned out that the angles between them are $50^o$, $60^o$ and $70^o$. Find the angles of the original triangle.
In a quadrilateral $ABCD$, we have $\angle DAB = 110^{\circ} , \angle ABC = 50^{\circ}$ and $\angle BCD = 70^{\circ}$ . Let $ M, N$ be the mid-points of $AB$ and $CD$ respectively. Suppose $P$ is a point on the segment $M N$ such that $\frac{AM}{CN} = \frac{MP}{PN}$ and $AP = CP$ . Find $\angle AP C$.
Let $M$ be the intersection of two diagonal, $AC$ and $BD$, of a rhombus $ABCD$, where angle $A<90^\circ$. Construct $O$ on segment $MC$ so that $OB<OC$ and let $t=\frac{MA}{MO}$, provided that $O \neq M$. Construct a circle that has $O$ as centre and goes through $B$ and $D$. Let the intersections between the circle and $AB$ be $B$ and $X$. Let the intersections between the circle and $BC$ be $B$ and $Y$. Let the intersections of $AC$ with $DX$ and $DY$ be $P$ and $Q$, respectively. Express $\frac{OQ}{OP}$ in terms of $t$.
You are given a quadrilateral $ABCD$. It is known that $\angle BAC = 30^o$, $\angle D = 150^o$ and, in addition, $AB = BD$. Prove that $AC$ is the bisector of angle $C$.
A circle centred at $I$ is tangent to the sides $BC, CA$, and $AB$ of an acute-angled triangle $ABC$ at $A_1, B_1$, and $C_1$, respectively. Let $K$ and $L$ be the incenters of the quadrilaterals $AB_1IC_1$ and $BA_1IC_1$, respectively. Let $CH$ be an altitude of triangle $ABC$. Let the internal angle bisectors of angles $AHC$ and $BHC$ meet the lines $A_1C_1$ and $B_1C_1$ at $P$ and $Q$, respectively. Prove that $Q$ is the orthocenter of the triangle $KLP$.
Kolmogorov Cup 2018, Major League, Day 3, Problem 1; A. Zaslavsky
Given a triangle $ABC$. The points $A$, $B$, $C$ divide the circumcircle $\Omega$ of the triangle $ABC$ into three arcs $BC$, $CA$, $AB$. Let $X$ be a variable point on the arc $AB$, and let $O_{1}$ and $O_{2}$ be the incenters of the triangles $CAX$ and $CBX$. Prove that the circumcircle of the triangle $XO_{1}O_{2}$ intersects the circle $\Omega$ in a fixed point.
Points $A$, $B$, $C$ and $D$ are chosen on a line in that order, with $AB$ and $CD$ greater than $BC$. Equilateral triangles $APB$, $BCQ$ and $CDR$ are constructed so that $P$, $Q$ and $R$ are on the same side with respect to $AD$. If $\angle PQR=120^\circ$, show that
\[\frac{1}{AB}+\frac{1}{CD}=\frac{1}{BC}.\]
Two different points $A$ and $B$ and a circle $\omega$ that passes through $A$ and $B$ are given. $P$ is a variable point on $\omega$ (different from $A$ and $B$). $M$ is a point such that $MP$ is the bisector of the angle $\angle{APB}$ ($M$ lies outside of $\omega$) and $MP=AP+BP$. Find the geometrical locus of $M$.