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$

In a triangle $ABC$, points $H,M,L$ are the feet of the altitude from $C$, the median from $A$, and the angle bisector from $B$, respectively. Show that if triangle $HML$ is equilateral, then so is triangle $ABC$.

Let $ABCD$ be an isosceles trapezoid with bases $AD> BC$. Let $X$ be the intersection of the bisectors of $\angle BAC$ and $BC$. Let $E$ be the intersection of$ DB$ with the parallel to the bisector of $\angle CBD$ through $X$ and let $F$ be the intersection of $DC$ with the parallel to the bisector of $\angle DCB$ through $X$. Show that quadrilateral $AEFD$ is cyclic.

In a triangle $ ABC$, the angle bisector at $ B$ intersects $ AC$ at $ D$ and the circumcircle again at $ E$. The circumcircle of the triangle $ DAE$ meets the segment $ AB$ again at $ F$. Prove that the triangles $ DBC$ and $ DBF$ are congruent.

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 a triangle. Let B' and C' denote the reflection of B and C in the internal angle bisector of angle A. Show that the triangles ABC and AB'C' have the same incenter.

If P is a point inside a convex quadrilateral $ABCD$, let the angle bisectors of $\angle APB$, $\angle BPC$, $\angle CPD$ and $\angle DPA$ meet $AB$, $BC$, $CD$ and $DA$ at $K$, $L$, $M$ and $N$ respectively. (a) Find a point $P$ such that $KLMN$ is a parallelogram. (b) Find the locus of all such points $P$. (S Tokarev)

A circle with center $O$ is inscribed in an angle. Let $A$ be the reflection of $O$ across one side of the angle. Tangents to the circle from $A$ intersect the other side of the angle at points $B$ and $C$. Prove that the circumcenter of triangle $ABC$ lies on the bisector of the original angle. (I.Sharygin)

A quadrilateral $ABCD$ is such that the sides $AB$ and $DC$ are parallel, and $|BC| =|AB| + |CD|$. Prove that the angle bisectors of the angles $\angle ABC$ and $\angle BCD$ intersect at right angles on the side $AD$.

In triangle ABC, the altitude, angle bisector and median from C divide the angle C into four equal angles. Find angle B.

Consider the set of 30 parabolas defined as follows: all parabolas have as focus the point (0,0) and the directrix lines have the form $y=ax+b$ with a and b integers such that $a\in \{-2,-1,0,1,2\}$ and $b\in \{-3,-2,-1,1,2,3\}$. No three of these parabolas have a common point. How many points in the plane are on two of these parabolas? ${ \textbf{(A)}\ 720\qquad\textbf{(B)}\ 760\qquad\textbf{(C)}\ 810\qquad\textbf{(D}}\ 840\qquad\textbf{(E)}\ 870 $

The sides $ a,b$ and the bisector of the included angle $ \gamma$ of a triangle are given. Determine necessary and sufficient conditions for such triangles to be constructible and show how to reconstruct the triangle.

Let $ABC$ be a triangle with $\angle BAC=117^\circ$. The angle bisector of $\angle ABC$ intersects side $AC$ at $D$. Suppose $\triangle ABD\sim\triangle ACB$. Compute the measure of $\angle ABC$, in degrees.

Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.) Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular. [i]Netherlands (Merlijn Staps)[/i]

If $BK$ is an angle bisector of $\angle ABC$ in triangle $ABC$. Find angles of triangle $ABC$ if $BK=KC=2AK$

Let $\beta$ be the length of the bisector of angle $B$, and $a', c'$ be the lengths of the segments into which this bisector divides the side $AC$ of the triangle $ABC$. Prove the relation $\beta^2 = ac-a'c'$ and derive from this the formula $\beta^2=ac-\frac{b^2ac}{(a+c)^2}$.

An equilateral triangle with side length $6$ has a square of side length $6$ attached to each of its edges as shown. The distance between the two farthest vertices of this figure (marked $A$ and $B$ in the figure) can be written as $m + \sqrt{n}$ where $m$ and $n$ are positive integers. Find $m + n$. [asy] draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle); draw((1,0)--(1+sqrt(3)/2,1/2)--(1/2+sqrt(3)/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2)); draw((0,0)--(-sqrt(3)/2,1/2)--(-sqrt(3)/2+1/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2)); dot((-sqrt(3)/2+1/2,1/2+sqrt(3)/2)); label("A", (-sqrt(3)/2+1/2,1/2+sqrt(3)/2), N); draw((1,0)--(1,-1)--(0,-1)--(0,0)); dot((1,-1)); label("B", (1,-1), SE); [/asy]

An acute-angled triangle $ABC$ is inscribed in a circle with center $O$. The bisector of $\angle A$ meets $BC$ at $D$, and the perpendicular to $AO$ through $D$ meets the segment $AC$ in a point $P$. Show that $AB = AP$.

Let $ABCD$ be a parallelogram. Let $M$ be a point on the side $AB$ and $N$ be a point on the side $BC$ such that the segments $AM$ and $CN$ have equal lengths and are non-zero. The lines $AN$ and $CM$ meet at $Q$. Prove that the line $DQ$ is the bisector of the angle $\measuredangle ADC$. [i]Alternative formulation.[/i] Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$. Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$.

Let $ABC$ be an isosceles triangle with $AC=BC$, let $M$ be the midpoint of its side $AC$, and let $Z$ be the line through $C$ perpendicular to $AB$. The circle through the points $B$, $C$, and $M$ intersects the line $Z$ at the points $C$ and $Q$. Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$.

Prove: In an acute triangle $ ABC$ angle bisector $ w_{\alpha},$ median $ s_b$ and the altitude $ h_c$ intersect in one point if $ w_{\alpha},$ side $ BC$ and the circle around foot of the altitude $ h_c$ have vertex $ A$ as a common point.

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 $PS =QS$.

In a triangle $\triangle{ABC}$ with all its sides of different length, $D$ is on the side $AC$, such that $BD$ is the angle bisector of $\sphericalangle{ABC}$. Let $E$ and $F$, respectively, be the feet of the perpendicular drawn from $A$ and $C$ to the line $BD$ and let $M$ be the point on $BC$ such that $DM$ is perpendicular to $BC$. Show that $\sphericalangle{EMD}=\sphericalangle{DMF}$.

Consider a triangle $ABC$ with $AB=AC$, and $D$ the foot of the altitude from the vertex $A$. The point $E$ lies on the side $AB$ such that $\angle ACE= \angle ECB=18^{\circ}$. If $AD=3$, find the length of the segment $CE$.

In a triangle $ABC$, the external bisector of $\angle BAC$ intersects the ray $BC$ at $D$. The feet of the perpendiculars from $B$ and $C$ to line $AD$ are $E$ and $F$, respectively and the foot of the perpendicular from $D$ to $AC$ is $G$. Show that $\angle DGE + \angle DGF = 180^{\circ}$.