Olympiad problems

2011 Portugal MO, 5

Let $[ABC]$ be a triangle, $D$ be the orthogonal projection of $B$ on the bisector of $\angle ACB$ and $E$ the orthogonal projection of $C$ on the bisector of $\angle ABC$ . Prove that $DE$ intersects the sides $[AB]$ and $[AC]$ at the touchpoints of the circle inscribed in the triangle $[ABC]$.

2015 Estonia Team Selection Test, 9

Tags: geometry , angle bisector , orthocenter , perpendicular

The orthocenter of an acute triangle $ABC$ is $H$. Let $K$ and $P$ be the midpoints of lines $BC$ and $AH$, respectively. The angle bisector drawn from the vertex $A$ of the triangle $ABC$ intersects with line $KP$ at $D$. Prove that $HD\perp AD$.

2018 Bosnia And Herzegovina - Regional Olympiad, 5

Tags: geometry , angle bisector , orthocenter , collinear

Let $H$ be an orhocenter of an acute triangle $ABC$ and $M$ midpoint of side $BC$. If $D$ and $E$ are foots of perpendicular of $H$ on internal and external angle bisector of angle $\angle BAC$, prove that $M$, $D$ and $E$ are collinear

2022 Yasinsky Geometry Olympiad, 6

Tags: geometry , angle bisector

Let $AD$, $BE$ and $CF$ be the diameters of the circle circumscribed around the acute angle triangle $ABC$. Point $N$ is the midpoint of the arc $CAD$, and point $M$ is the midpoint of arc $BAD$. Prove that the lines $EN$ and $MF$ intersect at the angle bisector of $\angle BAC$. (Matvii Kurskyi)

2023 Polish MO Finals, 2

Tags: incenter , angle bisector , geometry

Given an acute triangle $ABC$ with their incenter $I$. Point $X$ lies on $BC$ on the same side as $B$ wrt $AI$. Point $Y$ lies on the shorter arc $AB$ of the circumcircle $ABC$. It is given that $$\angle AIX = \angle XYA = 120^\circ.$$ Prove that $YI$ is the angle bisector of $XYA$.

2017 Regional Olympiad of Mexico Northeast, 2

Tags: geometry , angle bisector

Let $ABC$ be a triangle and let $N$ and $M$ be the midpoints of $AB$ and $CA$, respectively. Let $H$ be the foot of altitude from $A$. The circumcircle of $ABH$ intersects $MN$ at $P$, with $P$ and $M$ on the same side relative to $N$, and the circumcircle of $ACH$ intersects $MN$ at $Q$, with $Q$ and $N$ on the same side relative to $M$. $BP$ and $CQ$ intersect at $X$. Prove that $AX$ is the angle bisector of $\angle CAB$.

Swiss NMO - geometry, 2013.10

Tags: geometry , tangential quadrilateral , angle bisector , concurrency

Let $ABCD$ be a tangential quadrilateral with $BC> BA$. The point $P$ is on the segment $BC$, such that $BP = BA$ . Show that the bisector of $\angle BCD$, the perpendicular on line $BC$ through $P$ and the perpendicular on $BD$ through $A$, intersect at one point.

Ukraine Correspondence MO - geometry, 2015.8

Tags: geometry , equal angles , angle bisector , equilateral

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 $EDF$.

2010 Estonia Team Selection Test, 4

Tags: geometry , circumcircle , angle bisector , perpendicular , tangent

In an acute triangle $ABC$ the angle $C$ is greater than the angle $A$. Let $AE$ be a diameter of the circumcircle of the triangle. Let the intersection point of the ray $AC$ and the tangent of the circumcircle through the vertex $B$ be $K$. The perpendicular to $AE$ through $K$ intersects the circumcircle of the triangle $BCK$ for the second time at point $D$. Prove that $CE$ bisects the angle $BCD$.

1998 National Olympiad First Round, 17

Tags: geometry , angle bisector , similar triangles

In triangle $ ABC$, internal bisector of angle $ A$ intersects with $ BC$ at $ D$. Let $ E$ be a point on $ \left[CB\right.$ such that $ \left|DE\right|\equal{}\left|DB\right|\plus{}\left|BE\right|$. The circle through $ A$, $ D$, $ E$ intersects $ AB$ at $ F$, again. If $ \left|BE\right|\equal{}\left|AC\right|\equal{}7$, $ \left|AD\right|\equal{}2\sqrt{7}$ and $ \left|AB\right|\equal{}5$, then $ \left|BF\right|$ is $\textbf{(A)}\ \frac {7\sqrt {5} }{5} \qquad\textbf{(B)}\ \sqrt {7} \qquad\textbf{(C)}\ 2\sqrt {2} \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \sqrt {10}$

2007 All-Russian Olympiad, 3

Tags: circumcircle , asymptote , cyclic quadrilateral , angle bisector , geometry

$BB_{1}$ is a bisector of an acute triangle $ABC$. A perpendicular from $B_{1}$ to $BC$ meets a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$. A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$. $BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$ are collinear. [i]V. Astakhov[/i]

2022 South Africa National Olympiad, 4

Tags: geometry , circumcircle , angle bisector

Let $ABC$ be a triangle with $AB < AC$. A point $P$ on the circumcircle of $ABC$ (on the same side of $BC$ as $A$) is chosen in such a way that $BP = CP$. Let $BP$ and the angle bisector of $\angle BAC$ intersect at $Q$, and let the line through $Q$ and parallel to $BC$ intersect $AC$ at $R$. Prove that $BR = CR$.

2009 China Girls Math Olympiad, 2

Tags: geometry , circumcircle , incenter , angle bisector

Right triangle $ ABC,$ with $ \angle A\equal{}90^{\circ},$ is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA>EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC \equal{} \angle CAF.$ Segment $ BF$ meets $ \Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear.

2007 Nicolae Coculescu, 3

Tags: angle bisector , geometry

Let $ M,N $ be points on the segments $ AB,AC, $ respectively, of the triangle $ ABC. $ Also, let $ P,Q, $ be the midpoints of the segments $ MN,BC, $ respectively. Knowing that $ PQ $ is parallel to the bisector of $ \angle BAC , $ show that $ BM=CN. $ [i]Gheorghe Duță[/i]

2012 USA Team Selection Test, 2

Tags: trigonometry , reflection , cyclic quadrilateral , angle bisector , projective geometry , geometry

In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other.

2017 CHMMC (Fall), 7

Tags: geometry , angle bisector

Triangle $ABC$ has side lengths $AB=18$, $BC=36$, and $CA=24$. The circle $\Gamma$ passes through point $C$ and is tangent to segment $AB$ at point $A$. Let $X$, distinct from $C$, be the second intersection of $\Gamma$ with $BC$. Moreover, let $Y$ be the point on $\Gamma$ such that segment $AY$ is an angle bisector of $\angle XAC$. Suppose the length of segment $AY$ can be written in the form $AY=\frac{p\sqrt{r}}{q}$ where $p$, $q$, and $r$ are positive integers such that $gcd(p, q)=1$ and $r$ is square free. Find the value of $p+q+r$.

2004 District Olympiad, 2

Tags: geometry , angle bisector , equal segments , parallel

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.

1990 Canada National Olympiad, 3

Tags: geometry , incenter , cyclic quadrilateral , angle bisector

The feet of the perpendiculars from the intersection point of the diagonals of a convex cyclic quadrilateral to the sides form a quadrilateral $q$. Show that the sum of the lengths of each pair of opposite sides of $q$ is equal.

Ukrainian From Tasks to Tasks - geometry, 2016.3

Tags: geometry , angle bisector , concurrency

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]

2000 India Regional Mathematical Olympiad, 5

Tags: geometry , circumcircle , perpendicular bisector , angle bisector

The internal bisector of angle $A$ in a triangle $ABC$ with $AC > AB$ meets the circumcircle $\Gamma$ of the triangle in $D$. Join$D$ to the center $O$ of the circle $\Gamma$ and suppose that $DO$ meets $AC$ in $E$, possibly when extended. Given that $BE$ is perpendicular to $AD$, show that $AO$ is parallel to $BD$.

2012 Argentina Cono Sur TST, 5

Tags: angle bisector , excenter , geometry

Let $ABC$ be a triangle, and $K$ and $L$ be points on $AB$ such that $\angle ACK = \angle KCL = \angle LCB$. Let $M$ be a point in $BC$ such that $\angle MKC = \angle BKM$. If $ML$ is the angle bisector of $\angle KMB$, find $\angle MLC$.

2000 Junior Balkan Team Selection Tests - Moldova, 3

Tags: geometry , angle bisector , geometric inequality , isosceles

Let $ABC$ be a triangle with $AB = AC$ ¸ $\angle BAC = 100^o$ and $AD, BE$ angle bisectors. Prove that $2AD <BE + EA$

2010 IFYM, Sozopol, 3

Tags: circumcircle , congruent triangles , angle bisector , geometry

Let $ ABC$ is a triangle, let $ H$ is orthocenter of $ \triangle ABC$, let $ M$ is midpoint of $ BC$. Let $ (d)$ is a line perpendicular with $ HM$ at point $ H$. Let $ (d)$ meet $ AB, AC$ at $ E, F$ respectively. Prove that $ HE \equal{}HF$.

2010 Contests, 3

Tags: circumcircle , trigonometry , perpendicular bisector , angle bisector , geometry

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 $LN = NA$.

2010 Regional Olympiad of Mexico Center Zone, 1

Tags: angle bisector , circumcircle , geometry

In the acute triangle $ABC$, $\angle BAC$ is less than $\angle ACB $. Let $AD$ be a diameter of $\omega$, the circle circumscribed to said triangle. Let $E$ be the point of intersection of the ray $AC$ and the tangent to $\omega$ passing through $B$. The perpendicular to $AD$ that passes through $E$ intersects the circle circumscribed to the triangle $BCE$, again, at the point $F$. Show that $CD$ is an angle bisector of $\angle BCF$.