Found problems: 361
2020 Czech and Slovak Olympiad III A, 5
Given an isosceles triangle $ABC$ with base $BC$. Inside the side $BC$ is given a point $D$. Let $E, F$ be respectively points on the sides $AB, AC$ that $|\angle BED | = |\angle DF C| > 90^o$ . Prove that the circles circumscribed around the triangles $ABF$ and $AEC$ intersect on the line $AD$ at a point different from point $A$.
(Patrik Bak, Michal RolÃnek)
2006 Tournament of Towns, 4
In triangle $ABC$ let $X$ be some fixed point on bisector $AA'$ while point $B'$ be intersection of $BX$ and $AC$ and point $C'$ be intersection of $CX$ and $AB$. Let point $P$ be intersection of segments $A'B'$ and $CC'$ while point $Q$ be intersection of segments $A'C'$ and $BB'$. Prove Ï„hat $\angle PAC = \angle QAB$.
2020 Ukrainian Geometry Olympiad - April, 4
On the sides $AB$ and $AD$ of the square $ABCD$, the points $N$ and $P$ are selected respectively such that $NC=NP$. The point $Q$ is chosen on the segment $AN$ so that $\angle QPN = \angle NCB$. Prove that $2\angle BCQ = \angle AQP$.
2010 Dutch IMO TST, 4
Let $ABCD$ be a cyclic quadrilateral satisfying $\angle ABD = \angle DBC$. Let $E$ be the intersection of the diagonals $AC$ and $BD$. Let $M$ be the midpoint of $AE$, and $N$ be the midpoint of $DC$. Show that $MBCN$ is a cyclic quadrilateral.
Swiss NMO - geometry, 2012.3
The circles $k_1$ and $k_2$ intersect at points $D$ and $P$. The common tangent of the two circles on the side of $D$ touches $k_1$ at $A$ and $k_2$ at $B$. The straight line $AD$ intersects $k_2$ for a second time at $C$. Let $M$ be the center of the segment $BC$. Show that $ \angle DPM = \angle BDC$ .
Indonesia MO Shortlist - geometry, g5
Given a cyclic quadrilateral $ABCD$. Suppose $E, F, G, H$ are respectively the midpoint of the sides $AB, BC, CD, DA$. The line passing through $G$ and perpendicular on $AB$ intersects the line passing through $H$ and perpendicular on $BC$ at point $K$. Prove that $\angle EKF = \angle ABC$.
2020 Ukrainian Geometry Olympiad - April, 4
Inside triangle $ABC$, the point $P$ is chosen such that $\angle PAB = \angle PCB =\frac14 (\angle A+ \angle C)$. Let $BL$ be the bisector of $\vartriangle ABC$. Line $PL$ intersects the circumcircle of $\vartriangle APC$ at point $Q$. Prove that the line $QB$ is the bisector of $\angle AQC$.
Kyiv City MO Juniors Round2 2010+ geometry, 2013.8.3
Inside $\angle BAC = 45 {} ^ \circ$ the point $P$ is selected that the conditions $\angle APB = \angle APC = 45 {} ^ \circ $ are fulfilled. Let the points $M$ and $N$ be the projections of the point $P$ on the lines $AB$ and $AC$, respectively. Prove that $BC\parallel MN $.
(Serdyuk Nazar)
Geometry Mathley 2011-12, 13.1
Let $ABC$ be a triangle with no right angle, $E$ on the line $BC$ such that $\angle AEB = \angle BAC$ and $\Delta_A$ the perpendicular to $BC$ at $E$. Let the circle $\gamma$ with diameter $BC$ intersect $BA$ again at $D$. For each point $M$ on $\gamma$ ($M$ is distinct from $B$), the line $BM$ meets $\Delta_A$ at $M'$ and the line $AM$ meets $\gamma$ again at $M''$.
(a) Show that $p(A) = AM' \times DM''$ is independent of the chosen $M$.
(b) Keeping $B,C$ fixed, and let $A$ vary. Show that $\frac{p(A)}{d(A,\Delta_A)}$ is independent of $A$.
Michel Bataille
1995 Singapore Team Selection Test, 2
$ABC$ is a triangle with $\angle A > 90^o$ . On the side $BC$, two distinct points $P$ and $Q$ are chosen such that $\angle BAP = \angle PAQ$ and $BP \cdot CQ = BC \cdot PQ$. Calculate the size of $\angle PAC$.
2020 Dutch IMO TST, 1
In acute-angled triangle $ABC, I$ is the center of the inscribed circle and holds $| AC | + | AI | = | BC |$. Prove that $\angle BAC = 2 \angle ABC$.
2013 Grand Duchy of Lithuania, 2
Let $ABC$ be an isosceles triangle with $AB = AC$. The points $D, E$ and $F$ are taken on the sides $BC, CA$ and $AB$, respectively, so that $\angle F DE = \angle ABC$ and $FE$ is not parallel to $BC$. Prove that the line $BC$ is tangent to the circumcircle of $\vartriangle DEF$ if and only if $D$ is the midpoint of the side $BC$.
2015 Oral Moscow Geometry Olympiad, 1
Two trapezoid angles and diagonals are respectively equal. Is it true that such are the trapezoid equal?
2019 Saudi Arabia JBMO TST, 3
Consider a triangle $ABC$ and let $M$ be the midpoint of the side $BC$.
Suppose $\angle MAC = \angle ABC$ and $\angle BAM = 105^o$. Find the measure of $\angle ABC$.
Kyiv City MO Juniors 2003+ geometry, 2016.9.51
On the sides $AB$ and $AD$ of the square $ABCD$, the points $N$ and $P$ are selected, respectively, so that $PN = NC$, the point $Q$ Is a point on the segment $AN$ for which $\angle NCB = \angle QPN$. Prove that $\angle BCQ = \tfrac {1} {2} \angle PQA$.
1989 All Soviet Union Mathematical Olympiad, 492
$ABC$ is a triangle. $A' , B' , C'$ are points on the segments $BC, CA, AB$ respectively. $\angle B' A' C' = \angle A$ , $\frac{AC'}{C'B} = \frac{BA' }{A' C} = \frac{CB'}{B'A}$. Show that $ABC$ and $A'B'C'$ are similar.
2022 Dutch Mathematical Olympiad, 4
In triangle $ABC$, the point $D$ lies on segment $AB$ such that $CD$ is the angle bisector of angle $\angle C$. The perpendicular bisector of segment $CD$ intersects the line $AB$ in $E$. Suppose that $|BE| = 4$ and $|AB| = 5$.
(a) Prove that $\angle BAC = \angle BCE$.
(b) Prove that $2|AD| = |ED|$.
[asy]
unitsize(1 cm);
pair A, B, C, D, E;
A = (0,0);
B = (2,0);
C = (1.8,1.8);
D = extension(C, incenter(A,B,C), A, B);
E = extension((C + D)/2, (C + D)/2 + rotate(90)*(C - D), A, B);
draw((E + (0.5,0))--A--C--B);
draw(C--D);
draw(interp((C + D)/2,E,-0.3)--interp((C + D)/2,E,1.2));
dot("$A$", A, SW);
dot("$B$", B, S);
dot("$C$", C, N);
dot("$D$", D, S);
dot("$E$", E, S);
[/asy]
1994 Mexico National Olympiad, 3
$ABCD$ is a parallelogram. Take $E$ on the line $AB$ so that $BE = BC$ and $B$ lies between $A$ and $E$. Let the line through $C$ perpendicular to $BD$ and the line through $E$ perpendicular to $AB$ meet at $F$. Show that $\angle DAF = \angle BAF$.
Kyiv City MO Juniors Round2 2010+ geometry, 2015.789.4
In the acute triangle $ABC$ the side $BC> AB$, and the angle bisector $BL = AB$. On the segment $BL$ there is a point $M$, for which $\angle AML = \angle BCA$. Prove that $AM = LC$.
2010 IFYM, Sozopol, 8
In the trapezoid $ABCD, AB // CD$ and the diagonals intersect at $O$. The points $P, Q$ are on $AD, BC$ respectively such that $\angle AP B = \angle CP D$ and $\angle AQB = \angle CQD$. Show that $OP = OQ$.
2018 Yasinsky Geometry Olympiad, 1
In the triangle $ABC$, $AD$ is altitude, $M$ is the midpoint of $BC$. It is known that $\angle BAD = \angle DAM = \angle MAC$. Find the values of the angles of the triangle $ABC$
Russian TST 2019, P3
Let $\Omega$ be the circumcircle of an acute-angled triangle $ABC$. A point $D$ is chosen on the internal bisector of $\angle ACB$ so that the points $D$ and $C$ are separated by $AB$. A circle $\omega$ centered at $D$ is tangent to the segment $AB$ at $E$. The tangents to $\omega$ through $C$ meet the segment $AB$ at $K$ and $L$, where $K$ lies on the segment $AL$. A circle $\Omega_1$ is tangent to the segments $AL, CL$, and also to $\Omega$ at point $M$. Similarly, a circle $\Omega_2$ is tangent to the segments $BK, CK$, and also to $\Omega$ at point $N$. The lines $LM$ and $KN$ meet at $P$. Prove that $\angle KCE = \angle LCP$.
Poland
2018 Yasinsky Geometry Olympiad, 6
$AH$ is the altitude of the acute triangle $ABC$, $K$ and $L$ are the feet of the perpendiculars, from point $H$ on sides $AB$ and $AC$ respectively. Prove that the angles $BKC$ and $BLC$ are equal.
Kyiv City MO Juniors 2003+ geometry, 2019.8.3
In the triangle $ABC$ it is known that $2AC=AB$ and $\angle A = 2\angle B$. In this triangle draw the angle bisector $AL$, and mark point $M$, the midpoint of the side $AB$. It turned out that $CL = ML$. Prove that $\angle B= 30^o$.
(Hilko Danilo)
Kyiv City MO Seniors Round2 2010+ geometry, 2013.11.4
Let $ H $ be the intersection point of the altitudes $ AP $ and $ CQ $ of the acute-angled triangle $ ABC $. On its median $ BM $ marked points $ E $ and $ F $ so that $ \angle APE = \angle BAC $ and $ \angle CQF = \angle BCA $, and the point $ E $ lies inside the triangle $ APB $, and the point $ F $ lies inside the triangle $ CQB $. Prove that the lines $ AE $, $ CF $ and $ BH $ intersect at one point.
(Vyacheslav Yasinsky)