Found problems: 361
In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.
Let $\omega$ be the circumscribed circle of triangle $ABC$, and let $\omega'$ 'be the circle tangent to the side $BC$ and the extensions of the sides $AB$ and $AC$. The common tangents to the circles $\omega$ and $\omega'$ intersect the line $BC$ at points $D$ and $E$. Prove that $\angle BAD = \angle CAE$.
In the plane we have a line $r$ and two points $A$ and $B$ outside the line and in the same half plane. Determine a point $M$ on the line such that the angle of $r$ with $AM$ is double that of $r$ with $BM$. (Consider the smaller angle of two lines of the angles they form).
$AH$ is the altitude of the acute triangle $ABC$, $K$ and $L$ are the feet of the perpendiculars, from point $H$ on sides $AB$ and $AC$ respectively. Prove that the angles $BKC$ and $BLC$ are equal.
Let $ABCD$ be a cyclie quadrilateral, $\omega$ be it's circumcircle and $M$ be the midpoint of the arc $AB$ of $\omega$ which does not contain the vertices $C$ and $D$. The line that passes through $M$ and the intersection point of segments $AC$ and $BD$, intersects again $\omega$ in $N$. Let $P$ and $Q$ be points in the $CD$ segment such that $\angle AQD = \angle DAP$ and $\angle BPC = \angle CBQ$. Prove that the circumcircle of $NPQ$ and $\omega$ are tangent to each other.
Inside $\angle BAC = 45 {} ^ \circ$ the point $P$ is selected that the conditions $\angle APB = \angle APC = 45 {} ^ \circ $ are fulfilled. Let the points $M$ and $N$ be the projections of the point $P$ on the lines $AB$ and $AC$, respectively. Prove that $BC\parallel MN $.
(Serdyuk Nazar)
The altitudes of the acute-angled non-isosceles triangle $ ABC $ intersect at the point $ H $. On the segment $ C_1H $, where $ CC_1 $ is the altitude of the triangle, the point $ K $ is marked. Points $ L $ and $ M $ are the feet of perpendiculars from point $ K $ on straight lines $ AC $ and $ BC $, respectively. The lines $ AM $ and $ BL $ intersect at $ N $. Prove that $ \angle ANK = \angle HNL $.
In triangle $ABC$ let $X$ be some fixed point on bisector $AA'$ while point $B'$ be intersection of $BX$ and $AC$ and point $C'$ be intersection of $CX$ and $AB$. Let point $P$ be intersection of segments $A'B'$ and $CC'$ while point $Q$ be intersection of segments $A'C'$ and $BB'$. Prove τhat $\angle PAC = \angle QAB$.
In triangle $ABC$, the lengths of all sides are integers, $\angle B=2 \angle A$ and $\angle C> 90^o$. Find the smallest possible perimeter of this triangle.
The diagonals of a convex quadrilateral $ABCD$ intersect at $M$. The bisector of $\angle ACD$ intersects the ray $BA$ at $K$. Prove that if $MA\cdot MC + MA\cdot CD = MB \cdot MD $, then $\angle BKC = \angle BDC$
Find all values of $n$ for which there exists a convex cyclic non-regular polygon with $n$ vertices such that the measures of all its internal angles are equal.
Given a triangle $ABC$ with obtuse $\angle A$ and attitude $AH$ with $H \in BC$. Let $E,F$ on $CA$, $AB$ satisfying $\angle BEH = \angle C$ and $\angle CFH = \angle B$. Let $BE$ cut $CF$ at $D$. Prove that $DE = DF$.
Let $D$ be a point inside an acute triangle $ABC$, such that $\angle BAD = \angle DBC$ and $\angle DAC = \angle BCD$. Let $P$ be a point on the circumcircle of the triangle $ADB$. Suppose $P$ are itself outside the triangle $ABC$. A line through $P$ intersects the ray $BA$ in $X$ and ray $CA$ in $Y$, so that $\angle XPB = \angle PDB$. Show that $BY$ and $CX$ intersect on $AD$.
In each side of a regular polygon with $n$ sides, we choose a point different from the vertices and we obtain a new polygon of $n$ sides. For which values of $n$ can we obtain a polygon such that the internal angles are all equal but the polygon isn't regular?
Given a convex pentagon $ABCDE$ in which $\angle ABC = \angle AED = 90^o$, $\angle BAC= \angle DAE$. Let $K$ be the midpoint of the side $CD$, and $P$ the intersection point of lines $AD$ and $BK$, $Q$ be the intersection point of lines $AC$ and $EK$. Prove that $BQ = PE$.
In the triangle $ABC$, the heights $AA_1$ and $BB_1$ are drawn. On the side $AB$, points $M$ and $K$ are chosen so that $B_1K\parallel BC$ and $A_1 M\parallel AC$. Prove that the angle $AA_1K$ is equal to the angle $BB_1M$.
Let $ABCD$ be a cyclic quadrilateral with $\angle ADC = \angle DBA$. Furthermore, let $E$ be the projection of $A$ on $BD$. Show that $BC = DE - BE$ .
Point $M$ is the midpoint of the base $BC$ of trapezoid $ABCD$. On base $AD$, point $P$ is selected. Line $PM$ intersects line $DC$ at point $Q$, and the perpendicular from $P$ on the bases intersects line $BQ$ at point $K$. Prove that $\angle QBC = \angle KDA$.
The circles $ S_1 $ and $ S_2 $ intersect at points $ A $ and $ B $. Circle $ S_3 $ externally touches $ S_1 $ and $ S_2 $ at points $ C $ and $ D $ respectively. Let $ PQ $ be a chord cut by the line $ AB $ on circle $ S_3 $, and $ K $ be the midpoint of $ CD $. Prove that $ \angle PKC = \angle QKC $.
The circles $C_1$ and $C_2$ intersect at the distinct points $M$ and $N$. Points $A$ and $B$ belong respectively to the circles $C_1$ and $C_2$ so that the chords $[MA]$ and $[MB]$ are tangent at point $M$ to the circles $C_2$ and $C_1$, respectively. To prove it that the angles $\angle MNA$ and $\angle MNB$ are equal.
Determine the angles of a triangle in which the median, bisector and altitude, drawn from one vertex, divide this angle into four equal parts.
Let $ABCD$ be a parallelogram with center $O$ such that $\angle BAD <90^o$ and $\angle AOB> 90^o$. Consider points $A_1$ and $B_1$ on the rays $OA$ and $OB$ respectively, such that $A_1B_1$ is parallel to $AB$ and $\angle A_1B_1C = \frac12 \angle ABC$. Prove that $A_1D$ is perpendicular to $B_1C$.
Let $O$ be the center of the circle of the triangle $ABC$ with $AB \ne AC$. Furthermore, let $S$ and $T$ be points on the rays $AB$ and $AC$, such that $\angle ASO = \angle ACO$ and $\angle ATO = \angle ABO$. Show that $ST$ bisects the segment $BC$.
Let $ABC$ and $ABD$ be coplanar triangles with equal perimeters. The lines of support of the internal bisectrices of the angles $CAD$ and $CBD$ meet at $P$. Show that the angles $APC$ and $BPD$ are congruent.
Two circles $c$ and $c'$ with centers $O$ and $O'$ lie completely outside each other. Points $A, B$, and $C$ lie on the circle $c$ and points $A', B'$, and $C$ lie on the circle $c'$ so that segment $AB\parallel A'B'$, $BC \parallel B'C'$, and $\angle ABC = \angle A'B'C'$. The lines $AA', BB$', and $CC'$ are all different and intersect in one point $P$, which does not coincide with any of the vertices of the triangles $ABC$ or $A'B'C'$. Prove that $\angle AOB = \angle A'O'B'$.