$ABC$ is an acute-angled triangle. Let $D$ be the point on $BC$ such that $AD$ is the bisector of $\angle A$. Let $E, F$ be the feet of perpendiculars from $D$ to $AC,AB$ respectively. Suppose the lines $BE$ and $CF$ meet at $H$. The circumcircle of triangle $AFH$ meets $BE$ at $G$ (apart from $H$). Prove that the triangle constructed from $BG$, $GE$ and $BF$ is right-angled.

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]

In a triangle $ABC$, the angle bisector at $A,B,C$ meet the opposite sides at $A_1,B_1,C_1$, respectively. Prove that if the quadrilateral $BA_1B_1C_1$ is cyclic, then $$\frac{AC}{AB+BC}=\frac{AB}{AC+CB}+\frac{BC}{BA+AC}.$$

Angle bisector from vertex $A$ of acute triangle $ABC$ intersects side $BC$ in point $D$, and circumcircle of $ABC$ in point $E$ (different from $A$). Let $F$ and $G$ be foots of perpendiculars from point $D$ to sides $AB$ and $AC$. Prove that area of quadrilateral $AEFG$ is equal to the area of triangle $ABC$

Let the two tangents from a point $A$ outside a circle $k$ touch $k$ at $M$ and $N$. A line $p$ through $A$ intersects $k$ at $B$ and $C$, and $D$ is the midpoint of $MN$. Prove that $MN$ bisects the angle $BDC$

Let $ABC$ be a triangle, $I$ its incenter, and $D$ a point on the arc $BC$ of the circumcircle of $ABC$ not containing $A$. The bisector of the angle $\angle ADB$ intesects the segment $AB$ at $E$. The bisector of the angle $\angle CDA$ intesects the segment $AC$ at $F$. Prove that the points $E, F,I$ are collinear. (Malik Talbi)

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 $AQ=AR$.

Let $ABCD$ be a convex quadrileteral such that $m(\widehat{ABD})=40^\circ$, $m(\widehat{DBC})=70^\circ$, $m(\widehat{BDA})=80^\circ$, and $m(\widehat{BDC})=50^\circ$. What is $m(\widehat{CAD})$? $ \textbf{(A)}\ 25^\circ \qquad\textbf{(B)}\ 30^\circ \qquad\textbf{(C)}\ 35^\circ \qquad\textbf{(D)}\ 38^\circ \qquad\textbf{(E)}\ 40^\circ $

The triangle $ABC$ is drawn on the board such that $AB + AC = 2BC$. The bisectors $AL_1, BL_2, CL_3$ were drawn in this triangle, after which everything except the points $L_1, L_2, L_3$ was erased. Use a compass and a ruler to reconstruct triangle $ABC$.

Let \( n \geq 3 \) be a positive integer. In a convex polygon with \( n \) sides, all the internal bisectors of its \( n \) internal angles are drawn. Determine, as a function of \( n \), the smallest possible number of distinct lines determined by these bisectors.

Let $ ABCD$ be an isosceles trapezoid with $ AD \parallel BC$. Its diagonals $ AC$ and $ BD$ intersect at point $ M$. Points $ X$ and $ Y$ on the segment $ AB$ are such that $ AX \equal{} AM$, $ BY \equal{} BM$. Let $ Z$ be the midpoint of $ XY$ and $ N$ is the point of intersection of the segments $ XD$ and $ YC$. Prove that the line $ ZN$ is parallel to the bases of the trapezoid. [i]Author: A. Akopyan, A. Myakishev[/i]

In $\vartriangle ABC, \angle ABC = \angle ACB = 40^o$ . $BD$ bisects $\angle ABC$ , with $D$ located on $AC$. Prove that $BD + DA = BC$.

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 $\measuredangle ADC$.

In a triangle $ \vartriangle ABC $, let $ h_a, h_b $ and $ h_c $ the atlitudes. Let $ D $ be the point where the inner bisector of $ \angle BAC $ cuts to the side $ BC $ and $ d_a $ is the distance from the $ D $ point next to $ AB $. The distances $ d_b $ and $ d_c $ are similarly defined. Show that: $$ \dfrac {3} {2} \le \dfrac {d_a} {h_a} + \dfrac {d_b} {h_b} + \dfrac {d_c} {h_c} $$ For what kind of triangles does the equality hold?

In triangle $ABC$ the angular bisector from $A$ intersects the side $BC$ at the point $D$, and the angular bisector from $B$ intersects the side $AC$ at the point $E$. Furthermore $|AE| + |BD| = |AB|$. Prove that $\angle C = 60^o$ [img]https://1.bp.blogspot.com/-8ARqn8mLn24/XzP3P5319TI/AAAAAAAAMUQ/t71-imNuS18CSxTTLzYXpd806BlG5hXxACLcBGAsYHQ/s0/2018%2BMohr%2Bp5.png[/img]

In isosceles triangle $ABC$ ($AB = BC$) one draws the angle bisector $CD$. The perpendicular to $CD$ through the center of the circumcircle of $ABC$ intersects $BC$ at $E$. The parallel to $CD$ through $E$ meets $AB$ at $F$. Show that $BE$ = $FD$. [i]M. Sonkin[/i]

In triangle $ABC$, the angle bisectors are $BE$ and $CF$ (where $E, F$ are on the sides of the triangle), and their intersection point is $I$. Point $N$ lies on the circumcircle of $AEF$, and the angle $\angle IAN$ is right. The circumcircle of $AEF$ meets the line $NI$ a second time at the point $L$. Show that the circumcenter of $AIL$ lies on line $BC$.

Prove that the bisectors of the angles of a rectangle, extended to their mutual intersection, form a square.

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

$ABC$ is an acute-angled triangle. Let $D$ be the point on $BC$ such that $AD$ is the bisector of $\angle A$. Let $E, F$ be the feet of perpendiculars from $D$ to $AC,AB$ respectively. Suppose the lines $BE$ and $CF$ meet at $H$. The circumcircle of triangle $AFH$ meets $BE$ at $G$ (apart from $H$). Prove that the triangle constructed from $BG$, $GE$ and $BF$ is right-angled.

Let $ABC$ be an acute triangle such that $AB<AC$. Let $\Omega$ be its circumcircle, $O$ be its circumcenter, and $l$ be the internal angle bisector of $\angle BAC$. Suppose that the tangents to $\Omega$ at $B$ and $C$ intersect at $X$. Let $\omega$ be a circle whose center is $X$ and passes $B$, and $Y$ be the intersection of $l$ and $\omega$ which is chosen inside $\triangle ABC$. Let $D,E$ be the projections of $Y$ onto $AB,AC$, respectively. $OY$ meets $BC$ at $Z$. $ZD,ZE$ meet $l$ at $P,Q$, respectively. Prove that $BQ$ and $CP$ are parallel.

Given triangle $ABC$ with $|AC| > |BC|$. The point $M$ lies on the angle bisector of angle $C$, and $BM$ is perpendicular to the angle bisector. Prove that the area of triangle AMC is half of the area of triangle $ABC$. [img]https://cdn.artofproblemsolving.com/attachments/4/2/1b541b76ec4a9c052b8866acbfea9a0ce04b56.png[/img]

Let $ABC$ be a triangle and let $M,N,P$ be points on the line $BC$ such that $AM,AN,AP$ are the altitude, the angle bisector and the median of the triangle, respectively. It is known that $\frac{[AMP]}{[ABC]}=\frac{1}{4}$ and $\frac{[ANP]}{[ABC]}=1-\frac{\sqrt{3}}{2}$. Find the angles of triangle $ABC$.

Let $ABCD$ be a square, and $C$ the circle whose diameter is $AB.$ Let $Q$ be an arbitrary point on the segment $CD.$ We know that $QA$ meets $C$ on $E$ and $QB$ meets it on $F.$ Also $CF$ and $DE$ intersect in $M.$ show that $M$ belongs to $C.$

Let $ ABCD$ be a cyclic quadrilateral and $ O$ be the intersection of diagonal $ AC$ and $ BD$. The circumcircles of triangle $ ABO$ and the triangle $ CDO$ intersect at $ K$. Let $ L$ be a point such that the triangle $ BLC$ is similar to $ AKD$ (in that order). Prove that if $ BLCK$ is a convex quadrilateral, then it has an incircle.