Found problems: 361
2012 Balkan MO Shortlist, G6
Let $P$ and $Q$ be points inside a triangle $ABC$ such that $\angle PAC = \angle QAB$ and $\angle PBC = \angle QBA$. Let $D$ and $E$ be the feet of the perpendiculars from $P$ to the lines $BC$ and $AC$, and $F$ be the foot of perpendicular from $Q$ to the line $AB$. Let $M$ be intersection of the lines $DE$ and $AB$. Prove that $MP \perp CF$
Champions Tournament Seniors - geometry, 2001.4
Given a convex pentagon $ABCDE$ in which $\angle ABC = \angle AED = 90^o$, $\angle BAC= \angle DAE$. Let $K$ be the midpoint of the side $CD$, and $P$ the intersection point of lines $AD$ and $BK$, $Q$ be the intersection point of lines $AC$ and $EK$. Prove that $BQ = PE$.
2015 Romania Team Selection Tests, 1
Let $ABC$ and $ABD$ be coplanar triangles with equal perimeters. The lines of support of the internal bisectrices of the angles $CAD$ and $CBD$ meet at $P$. Show that the angles $APC$ and $BPD$ are congruent.
Kyiv City MO Juniors Round2 2010+ geometry, 2020.8.2
Given a convex quadrilateral $ABCD$, in which $\angle CBD = 90^o$, $\angle BCD =\angle CAD$ and $AD= 2BC$. Prove that $CA =CD$.
(Anton Trygub)
2017 Peru IMO TST, 11
Let $ABC$ be an acute and scalene of circumcircle $\Gamma$ and orthocenter $H$. Let $A_1,B_1,C_1$ be the second intersection points of the lines $AH, BH, CH$ with $\Gamma$, respectively. The lines that pass through $A_1,B_1,C_1$ and are parallel to $BC,CA, AB$ intersect again to $\Gamma$ at $A_2,B_2,C_2$, respectively. Let $M$ be the intersection point of $AC_2$ and $BC_1, N$ the intersection point of $BA_2$ and $CA_1$, and $P$ the intersection point of $CB_2$ and $AB_1$. Prove that $\angle MNB = \angle AMP$ .
Oliforum Contest III 2012, 3
Show that if equiangular hexagon has sides $a, b, c, d, e, f$ in order then $a - d = e - b = c - f$.
1999 Portugal MO, 6
In the triangle $[ABC], D$ is the midpoint of $[AB]$ and $E$ is the trisection point of $[BC]$ closer to $C$. If $\angle ADC= \angle BAE$ , find the measue of $\angle BAC$ .
1953 Kurschak Competition, 3
$ABCDEF$ is a convex hexagon with all its sides equal. Also $\angle A + \angle C + \angle E = \angle B + \angle D + \angle F$. Show that $\angle A = \angle D$, $\angle B = \angle E$ and $\angle C = \angle F$.
2005 Oral Moscow Geometry Olympiad, 2
A parallelogram of $ABCD$ is given. Line parallel to $AB$ intersects the bisectors of angles $A$ and $C$ at points $P$ and $Q$, respectively. Prove that the angles $ADP$ and $ABQ$ are equal.
(A. Hakobyan)
Champions Tournament Seniors - geometry, 2005.2
Given a triangle $ABC$, the line passing through the vertex $A$ symmetric to the median $AM$ wrt the line containing the bisector of the angle $\angle BAC$ intersects the circle circumscribed around the triangle $ABC$ at points $A$ and $K$. Let $L$ be the midpoint of the segment $AK$. Prove that $\angle BLC=2\angle BAC$.
Kyiv City MO Juniors 2003+ geometry, 2020.9.4
Let the point $D$ lie on the arc $AC$ of the circumcircle of the triangle $ABC$ ($AB < BC$), which does not contain the point $B$. On the side $AC$ are selected an arbitrary point $X$ and a point $X'$ for which $\angle ABX= \angle CBX'$. Prove that regardless of the choice of the point $X$, the circle circumscribed around $\vartriangle DXX'$, passes through a fixed point, which is different from point $D$.
(Nikolaev Arseniy)
1972 All Soviet Union Mathematical Olympiad, 159
Given a rectangle $ABCD$, points $M$ -- the midpoint of $[AD]$ side, $N$ -- the midpoint of $[BC]$ side. Let us take a point $P$ on the extension of the $[DC]$ segment over the point $D$. Let us denote the intersection point of lines $(PM)$ and $(AC)$ as $Q$. Prove that the $\angle QNM= \angle MNP$
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 τhat $\angle PAC = \angle QAB$.
1968 All Soviet Union Mathematical Olympiad, 094
Given an octagon with the equal angles. The lengths of all the sides are integers. Prove that the opposite sides are equal in pairs.
[u]alternate wording[/u]
Consider an octagon with equal angles and with rational sides. Prove that it has a center of symmetry.
2019 Romania National Olympiad, 3
Let $ABC$ be a triangle in which $\angle ABC = 45^o$ and $\angle BAC > 90^o$. Let $O$ be the midpoint of the side $[BC]$. Consider the point $M \in (AC)$ such that $\angle COM =\angle CAB$. Perpendicular from $M$ on $AC$ intersects line $AB$ at point $P$.
a) Find the measure of the angle $\angle BCP$.
b) Show that if $\angle BAC = 105^o$, then $PB = 2MO$.
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]
2007 Bulgarian Autumn Math Competition, Problem 11.3
In $\triangle ABC$ we have that $CC_{1}$ is an angle bisector. The points $P\in C_{1}B$, $Q\in BC$, $R\in AC$, $S\in AC_{1}$ satisfy $C_{1}P=PQ=QC$ and $CR=RS=SC_{1}$. Prove that $CC_{1}$ bisects $\angle SCP$.
2020 Ukrainian Geometry Olympiad - April, 5
Given a convex pentagon $ABCDE$, with $\angle BAC = \angle ABE = \angle DEA - 90^o$, $\angle BCA = \angle ADE$ and also $BC = ED$. Prove that $BCDE$ is parallelogram.
2021 Polish Junior MO First Round, 4
A convex quadrilateral $ABCD$ is given where $\angle DAB =\angle ABC = 120^o$ and $CD = 3$,$BC = 2$, $AB = 1$. Calculate the length of segment $AD$.
2022 China Northern MO, 1
As shown in the figure, given $\vartriangle ABC$ with $AB \perp AC$, $AB=BC$, $D$ is the midpoint of the side $AB$, $DF\perp DE$, $DE=DF$ and $BE \perp EC$. Prove that $\angle AFD= \angle CEF$.
[img]https://cdn.artofproblemsolving.com/attachments/9/2/f16a8c8c463874f3ccb333d91cdef913c34189.png[/img]
2001 239 Open Mathematical Olympiad, 5
The circles $ S_1 $ and $ S_2 $ intersect at points $ A $ and $ B $. Circle $ S_3 $ externally touches $ S_1 $ and $ S_2 $ at points $ C $ and $ D $ respectively. Let $ K $ be the midpoint of the chord cut by the line $ AB $ on circles $ S_3 $. Prove that $ \angle CKA = \angle DKA $.
2022 China Team Selection Test, 1
Given two circles $\omega_1$ and $\omega_2$ where $\omega_2$ is inside $\omega_1$. Show that there exists a point $P$ such that for any line $\ell$ not passing through $P$, if $\ell$ intersects circle $\omega_1$ at $A,B$ and $\ell$ intersects circle $\omega_2$ at $C,D$, where $A,C,D,B$ lie on $\ell$ in this order, then $\angle APC=\angle BPD$.
Ukraine Correspondence MO - geometry, 2007.9
In triangle $ABC$, the lengths of all sides are integers, $\angle B=2 \angle A$ and $\angle C> 90^o$. Find the smallest possible perimeter of this triangle.
I Soros Olympiad 1994-95 (Rus + Ukr), 9.7
Given an acute triangle $ABC$, in which $\angle BAC <30^o$. On sides $AC$ and $AB$ are taken respectively points $D$ and $E$ such that $\angle BDC=\angle BDE = \angle ADE = 60^o$. Prove that the centers of the circles. inscribed in triangles $ADE$, $BDE$ and $BCD$ do not lie on the same line.
2002 Switzerland Team Selection Test, 2
A point$ O$ inside a parallelogram $ABCD$ satisfies $\angle AOB + \angle COD = \pi$. Prove that $\angle CBO = \angle CDO$.