This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 361

2012 Dutch IMO TST, 4
Tags: geometry , right angle , equal angles , angle chasing
Let $\vartriangle ABC$ be a triangle. The angle bisector of $\angle CAB$ intersects$ BC$ at $L$. On the interior of line segments $AC$ and $AB$, two points, $M$ and $N$, respectively, are chosen in such a way that the lines $AL, BM$ and $CN$ are concurrent, and such that $\angle AMN = \angle ALB$. Prove that $\angle NML = 90^o$.

2011 Dutch IMO TST, 3
Tags: geometry , equal angles , angle chasing , circles
The circles $\Gamma_1$ and $\Gamma_2$ intersect at $D$ and $P$. The common tangent line of the two circles closest to point $D$ touches $\Gamma_1$ in A and $\Gamma_2$ in $B$. The line $AD$ intersects $\Gamma_2$ for the second time in $C$. Let $M$ be the midpoint of line segment $BC$. Prove that $\angle DPM = \angle BDC$.

1998 Czech and Slovak Match, 5
Tags: maximum value , trigonometry , equal angles , centroid , geometry
In a triangle $ABC$, $T$ is the centroid and $ \angle TAB = \angle ACT$. Find the maximum possible value of $sin \angle CAT +sin \angle CBT$.

Indonesia MO Shortlist - geometry, g8
Tags: integer , geometry , consecutive , equal angles
Prove that there is only one triangle whose sides are consecutive natural numbers and one of the angles is twice the other angle.

1996 Argentina National Olympiad, 4
Tags: perpendicular , geometry , parallelogram , equal angles , parallel
Let $ABCD$ be a parallelogram with center $O$ such that $\angle BAD <90^o$ and $\angle AOB> 90^o$. Consider points $A_1$ and $B_1$ on the rays $OA$ and $OB$ respectively, such that $A_1B_1$ is parallel to $AB$ and $\angle A_1B_1C = \frac12 \angle ABC$. Prove that $A_1D$ is perpendicular to $B_1C$.

Novosibirsk Oral Geo Oly VIII, 2016.1
Tags: geometry , equal angles
In the quadrilateral $ABCD$, angles $B$ and $C$ are equal to $120^o$, $AB = CD = 1$, $CB = 4$. Find the length $AD$.

2013 BMT Spring, 15
Tags: equal angles , geometry
Let $ABCD$ be a convex quadrilateral with $\angle ABD = \angle BCD$, $AD = 1000$, $BD = 2000$, $BC = 2001$, and $DC = 1999$. Point $E$ is chosen on segment $DB$ such that $\angle ABD = \angle ECD$. Find $AE$.

2018 Switzerland - Final Round, 4
Tags: geometry , equal angles , concurrency
Let $D$ be a point inside an acute triangle $ABC$, such that $\angle BAD = \angle DBC$ and $\angle DAC = \angle BCD$. Let $P$ be a point on the circumcircle of the triangle $ADB$. Suppose $P$ are itself outside the triangle $ABC$. A line through $P$ intersects the ray $BA$ in $X$ and ray $CA$ in $Y$, so that $\angle XPB = \angle PDB$. Show that $BY$ and $CX$ intersect on $AD$.

2005 Oral Moscow Geometry Olympiad, 2
Tags: geometry , parallelogram , equal angles , angle bisector
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)

2018 Iranian Geometry Olympiad, 5
Tags: geometry , parallelogram , equal angles , circumcircle
Suppose that $ABCD$ is a parallelogram such that $\angle DAC = 90^o$. Let $H$ be the foot of perpendicular from $A$ to $DC$, also let $P$ be a point along the line $AC$ such that the line $PD$ is tangent to the circumcircle of the triangle $ABD$. Prove that $\angle PBA = \angle DBH$. Proposed by Iman Maghsoudi

2010 Dutch IMO TST, 4
Tags: geometry , cyclic quadrilateral , equal angles , midpoint
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.

2011 Saudi Arabia BMO TST, 3
Tags: bisects segment , equal angles , pentagon , geometry
Let $ABCDE$ be a convex pentagon such that $\angle BAC = \angle CAD = \angle DAE$ and $\angle ABC = \angle ACD = \angle ADE$. Diagonals $BD$ and $CE$ meet at $P$. Prove that $AP$ bisects side $CD$.

2013 Saudi Arabia GMO TST, 2
Tags: combinatorial geometry , equal angles , polygon , angle , geometry , combinatorics
Find all values of $n$ for which there exists a convex cyclic non-regular polygon with $n$ vertices such that the measures of all its internal angles are equal.

2017-IMOC, G7
Tags: geometry , circumcircle , concyclic , equal angles
Given $\vartriangle ABC$ with circumcenter $O$. Let $D$ be a point satisfying $\angle ABD = \angle DCA$ and $M$ be the midpoint of $AD$. Suppose that $BM,CM$ intersect circle $(O)$ at another points $E, F$, respectively. Let $P$ be a point on $EF$ so that $AP$ is tangent to circle $(O)$. Prove that $A, P,M,O$ are concyclic. [img]https://2.bp.blogspot.com/-gSgUG6oywAU/XnSKTnH1yqI/AAAAAAAALdw/3NuPFuouCUMO_6KbydE-KIt6gCJ4OgWdACK4BGAYYCw/s320/imoc2017%2Bg7.png[/img]

Estonia Open Senior - geometry, 2018.2.5
Tags: reflection , equal angles , angle , geometry
Let $A'$ be the result of reflection of vertex $A$ of triangle ABC through line $BC$ and let $B'$ be the result of reflection of vertex $B$ through line $AC$. Given that $\angle BA' C = \angle BB'C$, can the largest angle of triangle $ABC$ be located: a) At vertex $A$, b) At vertex $B$, c) At vertex $C$?

2016 Auckland Mathematical Olympiad, 2
Tags: geometry , equal angles , square
In square $ABCD$, $\overline{AC}$ and $\overline{BD}$ meet at point $E$. Point $F$ is on $\overline{CD}$ and $\angle CAF = \angle FAD$. If $\overline{AF}$ meets $\overline{ED}$ at point $G$, and if $\overline{EG} = 24$ cm, then find the length of $\overline{CF}$.

2009 Puerto Rico Team Selection Test, 4
Tags: geometry , equal angles , parallelogram
The point $ M$ is chosen inside parallelogram $ ABCD$. Show that $ \angle MAB$ is congruent to $ \angle MCB$, if and only if $ \angle MBA$ and $ \angle MDA$ are congruent.

2011 Belarus Team Selection Test, 2
Tags: equal angles , geometry
Two different points $X,Y$ are marked on the side $AB$ of a triangle $ABC$ so that $\frac{AX \cdot BX}{CX^2}=\frac{AY \cdot BY}{CY^2}$ . Prove that $\angle ACX=\angle BCY$. I.Zhuk

2013 Junior Balkan Team Selection Tests - Romania, 4
Tags: geometry , orthocenter , equal angles , perpendicular , concurrency
Consider acute triangles $ABC$ and $BCD$, with $\angle BAC = \angle BDC$, such that $A$ and $D$ are on opposite sides of line $BC$. Denote by $E$ the foot of the perpendicular line to $AC$ through $B$ and by $F$ the foot of the perpendicular line to $BD$ through $C$. Let $H_1$ be the orthocenter of triangle $ABC$ and $H_2$ be the orthocenter of $BCD$. Show that lines $AD, EF$ and $H_1H_2$ are concurrent.

2001 Cuba MO, 2
Tags: geometry , equal angles
Let $M$ be the point of intersection of the diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$. Let$ K$ be the intersection point of the extension of side $AB$ (from $A$) with the bisector of the $\angle ACD$. If $MA \cdot MC + MA \cdot CD = MB\cdot MD$ , prove that $\angle BKC = \angle CDB$.

2014 Indonesia MO Shortlist, G5
Tags: midpoint , geometry , equal angles
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 $\angle EKF = \angle ABC$.

Kyiv City MO Juniors Round2 2010+ geometry, 2019.7.3
Tags: equal segments , equal angles , geometry
In the quadrilateral $ABCD$ it is known that $\angle ABD= \angle DBC$ and $AD= CD$. Let $DH$ be the altitude of $\vartriangle ABD$. Prove that $| BC - BH | = HA$. (Hilko Danilo)

2006 Belarusian National Olympiad, 5
Tags: convex quadrilateral , hyperbola , equal angles , analytic geometry , conic , geometry
A convex quadrilateral $ABCD$ Is placed on the Cartesian plane. Its vertices $A$ and $D$ belong to the negative branch of the graph of the hyperbola $y= 1/x$, the vertices $B$ and $C$ belong to the positive branch of the graph and point $B$ lies at the left of $C$, the segment $AC$ passes through the origin $(0,0)$. Prove that $\angle BAD = \angle BCD$. (I, Voronovich)

Kyiv City MO Juniors 2003+ geometry, 2018.9.5
Tags: geometry , bisects segment , equal angles
Given a triangle $ABC$, the perpendicular bisector of the side $AC$ intersects the angle bisector of the triangle $AK$ at the point $P$, $M$ - such a point that $\angle MAC = \angle PCB$, $\angle MPA = \angle CPK$, and points $M$ and $K$ lie on opposite sides of the line $AC$. Prove that the line $AK$ bisects the segment $BM$. (Anton Trygub)

2012 IMAR Test, 3
Tags: equal angles , geometry , reflection , angle bisector
Given a triangle $ABC$, let $D$ be a point different from $A$ on the external bisectrix $\ell$ of the angle $BAC$, and let $E$ be an interior point of the segment $AD$. Reflect $\ell$ in the internal bisectrices of the angles $BDC$ and $BEC$ to obtain two lines that meet at some point $F$. Show that the angles $ABD$ and $EBF$ are congruent.