Let $\triangle ABC$ be a triangle with $|AC|=2|AB|$ and let $O$ be its circumcenter. Let $D$ be the intersection of the bisector of $\angle A$ with $BC$. Let $E$ be the orthogonal projection of $O$ to $AD$ and let $F\ne D$ be the point on $AD$ satisfying $|CD|=|CF|$. Prove that $\angle EBF=\angle ECF$.

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]

Let $ABC$ be an acute-angled triangle in which $\angle ABC$ is the largest angle. Let $O$ be its circumcentre. The perpendicular bisectors of $BC$ and $AB$ meet $AC$ at $X$ and $Y$ respectively. The internal angle bisectors of $\angle AXB$ and $\angle BYC$ meet $AB$ and $BC$ at $D$ and $E$ respectively. Prove that $BO$ is perpendicular to $AC$ if $DE$ is parallel to $AC$.

Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.

In triangle $ABC, \angle BAC = 60^o, \angle ACB = 90^o$ and $D$ is on $BC$. If $AD$ bisects $\angle BAC$ and $CD = 3$ cm, calculate $DB$ .

Lei it \(ABCD\) be a non-cyclical, convex quadrilateral, with no parallel sides. The lines \(AB\) and \(CD\) meet in \(E\). Let it \(M \not= E\) be the intersection of circumcircles of \(ADE\) and \(BCE\). The internal angle bisectors of \(ABCD\) form an convex, cyclical quadrilateral with circumcenter \(I\). The external angle bisectors of \(ABCD\) form an convex, cyclical quadrilateral with circumcenter \(J\). Show that \(I,J,M\) are colinear.

$ABCD$ is a trapezoid with $AB$ parallel to $CD$. The external bisectors of the angles at $ B$ and $C$ intersect at $ P$. The external bisectors of the angles at $ A$ and $D$ intersect at $Q$. Show that the length of $PQ$ is equal to half the perimeter of the trapezoid $ABCD$.

On the sides $BC, AC$ and $AB$ of the equilateral triangle $ABC$ mark the points $D, E$ and $F$ so that $\angle AEF = \angle FDB$ and $\angle AFE = \angle EDC$. Prove that $DA$ is the bisector of the angle $EDF$.

Let $ABC$ be an equilateral triangle. $M$ is the midpoint of segment $AB$ and $N$ is the midpoint of segment $BC$. Let $P$ be the point outside $ABC$ such that the triangle $ACP$ is isosceles and right in $P$. $PM$ and $AN$ are cut in $I$. Prove that $CI$ is the bisector of the angle $MCA$ .

Let $ABCD$ be a parallelogram. The interior angle bisector of $\angle ADC$ intersects the line $BC$ in $E$, and the perpendicular bisector of the side $AD$ intersects the line $DE$ in $M$. Let $F= AM \cap BC$. Prove that: a) $DE=AF$; b) $AD\cdot AB = DE\cdot DM$. [i]Daniela and Marius Lobaza, Timisoara[/i]

Given a triangle $ABC$ in which $|AB|>|AC|$. Let $P$ be the midpoint of the side $BC$, and $S$ the point in which the angle bisector of $\angle BAC$ intersects that side. The parallel with the line $AS$ through the point $P$ intersects lines $AB$ and $AC$ at points $X$ and $Y$ respectively . Let $Z$ be the point be such that $Y$ is the midpoint of the length $XZ$ and let the lines $BY$ and $CZ$ intersect at point $D$. Prove that the angle bisector of $\angle BDC$ is parallel to the lines $AS$.

The quadrilateral $ ABCD$ inscribed in a circle wich has diameter $ BD$. Let $ A',B'$ are symmetric to $ A,B$ with respect to the line $ BD$ and $ AC$ respectively. If $ A'C \cap BD \equal{} P$ and $ AC\cap B'D \equal{} Q$ then prove that $ PQ \perp AC$

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 $\angle APD$.

Let $ABCD$ be a convex quadrilateral. The bisector of the angle $ACD$ intersects $BD$ at point $E$. It is known that $\angle CAD = \angle BCE= 90^o$. Prove that the $AC$ is the bisector of the angle $BAE$ . (Nikolay Nikolay)

In fig. the bisectors of the angles $\angle DAC$, $ \angle EBD$, $\angle ACE$, $\angle BDA$ and $\angle CEB$ intersect at one point. Prove that the bisectors of the angles $\angle TPQ$, $\angle PQR$, $\angle QRS$, $\angle RST$ and $\angle STP$ also intersect at one point. [img]https://cdn.artofproblemsolving.com/attachments/6/e/870e4f20bc7fdcb37534f04541c45b1cd5034a.png[/img]