Let $ABCDEF$ be a regular hexagon and $M\in (DE)$, $N\in(CD)$ such that $m (\widehat {AMN}) = 90^\circ$ and $AN = CM \sqrt {2}$. Find the value of $\frac{DM}{ME}$.

Let $ABCD$ be a convex cyclic quadrilateral. Prove that: $a)$ the number of points on the circumcircle of $ABCD$, like $M$, such that $\frac{MA}{MB}=\frac{MD}{MC}$ is $4$. $b)$ The diagonals of the quadrilateral which is made with these points are perpendicular to each other.

An arbitrary point $M$ is selected in the interior of the segment $AB$. The square $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with segments $AM$ and $MB$ as their respective bases. The circles circumscribed about these squares, with centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$. a) Prove that $N$ and $N'$ coincide; b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$; c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.

The bisector of angle $A$ in parallelogram $ABCD$ intersects side $BC$ at $M$ and the bisector of $\angle AMC$ passes through point $D$. Find angles of the parallelogram if it is known that $\angle MDC = 45^o$. [img]https://cdn.artofproblemsolving.com/attachments/e/7/7cfb22f0c26fe39aa3da3898e181ae013a0586.png[/img]

Angle bisectors $AD$ and $BE$ are drawn in triangle $ABC$. It turned out that $DE$ is the bisector of triangle $ADC$. Find the angle $BAC$.

Two not necessarily equal non-intersecting wooden disks, one gray and one black, are glued to a plane. An infinite angle with one gray side and one black side can be moved along the plane so that the disks remain outside the angle, while the colored sides of the angle are tangent to the disks of the same color (the tangency points are not the vertices). Prove that it is possible to draw a ray in the angle, starting from the vertex of the angle and such that no matter how the angle is positioned, the ray passes through some fixed point of the plane. (Egor Bakaev, Ilya Bogdanov, Pavel Kozhevnikov, Vladimir Rastorguev) (Junior version [url=https://artofproblemsolving.com/community/c6h2094701p15140671]here[/url]) [hide=note]There was a mistake in the text of the problem 3, we publish here the correct version. The solutions were estimated according to the text published originally.[/hide]

Sides of a triangle are equal to $3$, $4$ and $5$. Each side is extended until it intersects the bisector of the external angle to the angle opposite to it. Three such points are obtained in all. Prove that one of the three points we get is the midpoint of the segment joining the other two points. (V. Prasolov)

the edges of triangle $ABC$ are $a,b,c$ respectively,$b<c$,$AD$ is the bisector of $\angle A$,point $D$ is on segment $BC$. (1)find the property $\angle A$,$\angle B$,$\angle C$ have,so that there exists point $E,F$ on $AB,AC$ satisfy $BE=CF$,and $\angle NDE=\angle CDF$ (2)when such $E,F$ exist,express $BE$ with $a,b,c$

Let $ ABC$ be a triangle, and let the angle bisectors of its angles $ CAB$ and $ ABC$ meet the sides $ BC$ and $ CA$ at the points $ D$ and $ F$, respectively. The lines $ AD$ and $ BF$ meet the line through the point $ C$ parallel to $ AB$ at the points $ E$ and $ G$ respectively, and we have $ FG \equal{} DE$. Prove that $ CA \equal{} CB$. [i]Original formulation:[/i] Let $ ABC$ be a triangle and $ L$ the line through $ C$ parallel to the side $ AB.$ Let the internal bisector of the angle at $ A$ meet the side $ BC$ at $ D$ and the line $ L$ at $ E$ and let the internal bisector of the angle at $ B$ meet the side $ AC$ at $ F$ and the line $ L$ at $ G.$ If $ GF \equal{} DE,$ prove that $ AC \equal{} BC.$

Let $ABC$ be a triangle and $D$ a point on the side $BC$. The angle bisectors of $\angle ADB ,\angle ADC$ intersect $AB ,AC$ at points $M ,N$ respectively. The angle bisectors of $\angle ABD , \angle ACD$ intersects $DM , DN$ at points $K , L$ respectively. Prove that $AM = AN$ if and only if $MN$ and $KL$ are parallel.

Let $ABCD$ be a convex quadrilateral inscibed in circle $(O)$ such that $DB = DA + DC$. The point $P$ lies on the ray $AC$ such that $AP = BC$. The point $E$ is on $(O)$ such that $BE \perp AD$. Prove that $DP$ is parallel to the angle bisector of $\angle BEC$.

Two circles $(O)$ and $(O')$ intersect at $A$ and $B$. Take two points $P,Q$ on $(O)$ and $(O')$, respectively, such that $AP=AQ$. The line $PQ$ intersects $(O)$ and $(O')$ respectively at $M,N$. Let $E,F$ respectively be the centers of the two arcs $BP$ and $BQ$ (which don't contains $A$). Prove that $MNEF$ is a cyclic quadrilateral.

Points $B,C$ vary on two fixed rays emanating from point $A$ such that $AB+AC$ is constant. Show that there is a point $D$, other than $A$, such that the circumcircle of triangle $ABC$ passes through $D$ for all possible choices of $B, C$.

Let $ I $ be the incenter of triangle $ ABC $. The incircle touches $ BC, CA, AB$ at points $ P, Q, R $. A circle passing through $ B , C $ is tangent to the circle $I$ at point $ X $, a circle passing through $ C , A $ is tangent to the circle $I$ at point $ Y $, and a circle passing through $ A , B $ is tangent to the circle $I$ at point $ Z $, respectively. Prove that three lines $ PX, QY, RZ $ are concurrent.

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.

Let $ABCD$ be a cyclic quadrilateral, $E = AC \cap BD$, $F = AD \cap BC$. The bisectors of angles $AFB$ and $AEB$ meet $CD$ at points $X, Y$ . Prove that $A, B, X, Y$ are concyclic.

Let $ABCD$ be a trapezium (trapezoid) with $AD$ parallel to $BC$ and $J$ be the intersection of the diagonals $AC$ and $BD$. Point $P$ a chosen on the side $BC$ such that the distance from $C$ to the line $AP$ is equal to the distance from $B$ to the line $DP$. [i]The following three questions 1, 2 and 3 are independent, so that a condition in one question does not apply in another question.[/i] 1.Suppose that $Area( \vartriangle AJB) =6$ and that $Area(\vartriangle BJC) = 9$. Determine $Area(\vartriangle APD)$. 2. Find all points $Q$ on the plane of the trapezium such that $Area(\vartriangle AQB) = Area(\vartriangle DQC)$. 3. Prove that $PJ$ is the angle bisector of $\angle APD$.

The circle inscribed in triangle $ABC$ touches $AC$ at point $F$. The perpendicular from point $F$ on $BC$ intersects the bisector of angle $C$ at point $N$. Prove that segment $FN$ is equal to the radius of the circle inscribed in triangle $ABC$. (Oleksii Karliuchenko)

A half-circle of diameter $EF$ is placed on the side $BC$ of a triangle $ABC$ and it is tangent to the sides $AB$ and $AC$ in the points $Q$ and $P$ respectively. Prove that the intersection point $K$ between the lines $EP$ and $FQ$ lies on the altitude from $A$ of the triangle $ABC$. [i]Albania[/i]

Let there be a convex quadrilateral $ABCD$. The angle bisector of $\angle A$ meets the angle bisector of $\angle B$, the angle bisector of $\angle D$ at $P, Q$ respectively. The angle bisector of $\angle C$ meets the angle bisector of $\angle D$, the angle bisector of $\angle B$ at $R, S$ respectively. $P, Q, R, S$ are all distinct points. $PR$ and $QS$ meets perpendicularly at point $Z$. Denote $l_A, l_B, l_C, l_D$ as the exterior angle bisectors of $\angle A, \angle B, \angle C, \angle D$. Denote $E = l_A \cap l_B$, $F= l_B \cap l_C$, $G = l_C \cap l_D$, and $H= l_D \cap l_A$. Let $K, L, M, N$ be the midpoints of $FG, GH, HE, EF$ respectively. Prove that the area of quadrilateral $KLMN$ is equal to $ZM \cdot ZK + ZL \cdot ZN$.

Cyclic quadrilateral $ABCD$ has circumcircle $(O)$. Points $M$ and $N$ are the midpoints of $BC$ and $CD$, and $E$ and $F$ lie on $AB$ and $AD$ respectively such that $EF$ passes through $O$ and $EO=OF$. Let $EN$ meet $FM$ at $P$. Denote $S$ as the circumcenter of $\triangle PEF$. Line $PO$ intersects $AD$ and $BA$ at $Q$ and $R$ respectively. Suppose $OSPC$ is a parallelogram. Prove that $AQ=AR$.

Angle bisectors $AA', BB'$and $CC'$ are drawn in triangle $ABC$ with angle $\angle B= 120^o$. Find $\angle A'B'C'$.

A rectangle $ABCD$ is given. Segment $DK$ is equal to $BD$ and lies on the half-line $DC$. $M$ is the midpoint of $BK$. Prove that $AM$ is the angle bisector of $\angle BAC$. [i]Proposed by S. Berlov[/i]

In an acute-angled triangle $ABC$, point $I$ is the incenter, $H$ is the orthocenter, $O$ is the center of the circumscribed circle, $T$ and $K$ are the touchpoints of the $A$-excircle and incircle with side $BC$ respectively. It turned out that the segment $TI$ is passing through the point $O$. Prove that $HK$ is the angle bisector of $\angle BHC$. (Matvii Kurskyi)

In triangle $ABC$, the angle at vertex $B$ is $120^o$. Let $A_1, B_1, C_1$ be points on the sides $BC, CA, AB$ respectively such that $AA_1, BB_1, CC_1$ are bisectors of the angles of triangle $ABC$. Determine the angle $\angle A_1B_1C_1$.