Found problems: 68
Given an acute-angled triangle $ABC$ in which $|AB| <|AC|$. Point $D$ is the midpoint of the shorter arc $BC$ of its circumcircle. The point $I$ is the center of its incircle, and the point $J$ is symmetric point of $I$ wrt line $BC$. The line $DJ$ intersects the circumcircle of the triangle $ABC$ at the point $E$ belonging to the arc $AB$. Prove that $|AI |= |IE|$.
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
Let $ABC$ be an acute triangle and let $X, Y , Z$ denote the midpoints of the shorter arcs $BC, CA, AB$ of its circumcircle, respectively. Let $M$ be an arbitrary point on side $BC$. The line through $M$, parallel to the inner angular bisector of $\angle CBA$ meets the outer angular bisector of $\angle BCA$ at point $N$. The line through $M$, parallel to the inner angular bisector of $\angle BCA$ meets the outer angular bisector of $\angle CBA$ at point $P$. Prove that lines $XM, Y N, ZP$ pass through a single point.
Let $I$ be the incenter of the triangle $ABC$ ($AB < BC$). Let $M$ be the midpoint of $AC$, and let $N$ be the midpoint of the arc $AC$ of the circumcircle of $ABC$ which contains $B$. Prove that $\angle IMA = \angle INB$.
Let $ABC$ a triangle, $I$ the incenter, $D$ the contact point of the incircle with the side $BC$ and $E$ the foot of the bisector of the angle $A$. If $M$ is the midpoint of the arc $BC$ which contains the point $A$ of the circumcircle of the triangle $ABC$ and $\{F\} = DI \cap AM$, prove that $MI$ passes through the midpoint of $[EF]$.
Let $\triangle ABC$ have incenter $I$. The line $CI$ intersects the circumcircle of $\triangle ABC$ for the second time at $L$, and $CI=2IL$. Points $M$ and $N$ lie on the segment $AB$, such that $\angle AIM =\angle BIN = 90^{\circ}$. Prove that $AB=2MN$.
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
Let ${ABC}$ be a triangle inscribed in a circle. Point ${P}$ is the center of the arc ${BAC}$. The circle with the diameter ${CP}$ intersects the angle bisector of angle ${\angle BAC}$ at points ${K, L}$ ${(|AK| <|AL|)}$. Point ${M}$ is the reflection of ${L}$ with respect to line ${BC}$. Prove that the circumcircle of the triangle ${BKM}$ passes through the center of the segment ${BC}$ .
In acute triangle $ABC$, $AB>AC$. Let $I$ the incenter, $\Omega$ the circumcircle of triangle $ABC$, and $D$ the foot of perpendicular from $A$ to $BC$. $AI$ intersects $\Omega$ at point $M(\neq A)$, and the line which passes $M$ and perpendicular to $AM$ intersects $AD$ at point $E$. Now let $F$ the foot of perpendicular from $I$ to $AD$.
Prove that $ID\cdot AM=IE\cdot AF$.
In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is the midpoint of arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear.
The circle is divided into four non-overlapping gaps $ AB $, $ BC $, $ CD $ and $ DA $. Prove that the segment joining the midpoints of the arcs $AB$ and $CD$ is perpendicular to the segment joining the midpoints of the arcs $BC$ and $DA$.
Let $ABC$ be an acute scalene triangle. Let $D, E$ be points on segments $AB, AC$ respectively, such that $BD=CE$. Prove that the nine-point centers of $ADE$, $ACD$, $ABC$, $AEB$ form a rhombus.
[i]Proposed by Malay Mahajan and Siddharth Choppara[/i]
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
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 a non-isosceles triangle $ABC$ the centre of the incircle is denoted by $I$. The other intersection point of the angle bisector of $\angle BAC$ and the circumcircle of $\vartriangle ABC$ is $D$. The line through $I$ perpendicular to $AD$ intersects $BC$ in $F$. The midpoint of the circle arc $BC$ on which $A$ lies, is denoted by $M$. The other intersection point of the line $MI$ and the circle through $B, I$ and $C$, is denoted by $N$. Prove that $FN$ is tangent to the circle through $B, I$ and $C$.
Consider an arc $AB$ of a circle $C$ and a point $P$ variable in that arc $AB$. Let $D$ be the midpoint of the arc $AP$ that doeas not contain $B$ and let $E$ be the midpoint of the arc $BP$ that does not contain $A$. Let $C_1$ be the circle with center $D$ passing through $A$ and $C_2$ be the circle with center $E$ passing through $B.$ Prove that the line that contains the intersection points of $C_1$ and $C_2$ passes through a fixed point.
Given a triangle $ABC$ ($AB> AC$) and a circle circumscribed around it. Construct with a compass and a ruler the midpoint of the arc $BC$ (not containing vertex $A$), with no more than two lines (straight or circles).
Let $ABC$ be a triangle with circumcirle $\Gamma$, and let $M$ and $N$ be the respective midpoints of the minor arcs $AB$ and $AC$ of $\Gamma$. Let $P$ and $Q$ be points such that $AB=BP$, $AC=CQ$, and $P$, $B$, $C$, $Q$ lie on $BC$ in that order. Prove that $PM$ and $QN$ meet at a point on $\Gamma$.
[i]Proposed by Victor Domínguez[/i]
Given is the acute triangle $ABC$ with the intersection of altitudes $H$. The angle bisector of angle $BHC$ intersects side $BC$ at point $D$. Mark $E$ and $F$ the symmetrics of the point $D$ wrt lines $AB$ and $AC$. Prove that the circle circumscribed around the triangle $AEF$ passes through the midpoint of the arc $BAC$
Let $M $be the point of intersection of the diagonals of the inscribed quadrilateral $ABCD$. Prove that if $AB = AM,$ then a line passing through $M$ perpendicular to $AD$ passes through the midpoint of the arc $BC$.
Given two circles $\Gamma_1$ and $\Gamma_2$ which intersect at points $A$ and $B$. A line through $A$ intersects $\Gamma_1$ and $\Gamma_2$ at points $C$ and $D$, respectively. Let $M$ be the midpoint of arc $BC$ in $\Gamma_1$ ,which does not contains $A$, and $N$ is the midpoint of the arc $BD$ in $\Gamma_2$, which does not contain $A$. If $K$ is the midpoint of $CD$, prove that $\angle MKN = 90^o.$
Let $ABC$ be an acute triangle. Let $M_A, M_B$ and $M_C$ be the midpoints of sides $BC,CA$, respectively $AB$. Let $M'_A , M'_B$ and $M'_C$ be the the midpoints of the arcs $BC, CA$ and $AB$ respectively of the circumscriberd circle of triangle $ABC$. Let $P_A$ be the intersection of the straight line $M_BM_C$ and the perpendicular to $M'_BM'_C$ through $A$. Define $P_B$ and $P_C$ similarly. Show that the straight line $M_AP_A, M_BP_B$ and $M_CP_C$ intersect at one point.
A circumcircle of triangle $ABC$ meets the sides $AD$ and $CD$ of a parallelogram $ABCD$ at points $K$ and $L$ respectively. Let $M$ be the midpoint of arc $KL$ not containing $B$. Prove that $DM \perp AC$.
by E.Bakaev
Let $ABC$ be an acute triangle. Let $M_A, M_B$ and $M_C$ be the midpoints of sides $BC,CA$, respectively $AB$. Let $M'_A , M'_B$ and $M'_C$ be the the midpoints of the arcs $BC, CA$ and $AB$ respectively of the circumscriberd circle of triangle $ABC$. Let $P_A$ be the intersection of the straight line $M_BM_C$ and the perpendicular to $M'_BM'_C$ through $A$. Define $P_B$ and $P_C$ similarly. Show that the straight line $M_AP_A, M_BP_B$ and $M_CP_C$ intersect at one point.