Olympiad problems

2016 Sharygin Geometry Olympiad, P1

A trapezoid $ABCD$ with bases $AD$ and $BC$ is such that $AB = BD$. Let $M$ be the midpoint of $DC$. Prove that $\angle MBC$ = $\angle BCA$.

Kyiv City MO Juniors Round2 2010+ geometry, 2019.7.3

Tags: equal segments , geometry , equal angles

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)

2019 Saint Petersburg Mathematical Olympiad, 4

Tags: geometry , centroid , equal angles , circumcircle

Given a convex quadrilateral $ABCD$. The medians of the triangle $ABC$ intersect at point $M$, and the medians of the triangle $ACD$ at point$ N$. The circle, circumscibed around the triangle $ACM$, intersects the segment $BD$ at the point $K$ lying inside the triangle $AMB$ . It is known that $\angle MAN = \angle ANC = 90^o$. Prove that $\angle AKD = \angle MKC$.

2006 Cuba MO, 9

Tags: equal angles , perpendicular , geometry

In the cyclic quadrilateral $ABCD$, the diagonals $AC$ and $BD$ intersect at $P$. Let $O$ be the center of the circumcircle $ABCD$, and $E$ a point of the extension of $OC$ beyond $C$. A parallel line to $CD$ is drawn through $E$ that cuts the extension of $OD$ beyonf $D$ at $F$. Let $Q$ be a point interior to $ABCD$, such that $\angle AFQ = \angle BEQ$ and $\angle FAQ = \angle EBQ$. Prove that $PQ \perp CD$.

2019 Peru MO (ONEM), 3

Tags: concurrency , equal angles , trapezoid , geometry

In the trapezoid $ABCD$ , the base $AB$ is smaller than the $CD$ base. The point $K$ is chosen such that $AK$ is parallel to BC and $BK$ is parallel to $AD$. The points $P$ and $Q$ are chosen on the $AK$ and $BK$ rays respectively, such that $\angle ADP = \angle BCK$ and $\angle BCQ = \angle ADK$. (a) Show that the lines $AD, BC$ and $PQ$ go through the same point. (b) Assuming that the circumscribed circumferences of the $APD$ and $BCQ$ triangles intersect at two points, show that one of those points belongs to the line $PQ$.

2011 Romania National Olympiad, 3

Tags: midpoint , angle bisector , equal angles , geometry

In the convex quadrilateral $ABCD$ we have that $\angle BCD = \angle ADC \ge 90 ^o$. The bisectors of $\angle BAD$ and $\angle ABC$ intersect in $M$. Prove that if $M \in CD$, then $M$ is the middle of $CD$.

2010 IFYM, Sozopol, 8

Tags: trapezoid , equal angles , geometry , equal segments

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 $OP = OQ$.

2016 Portugal MO, 4

Tags: geometry , equal angles , tangent , angle , parallelogram

Let $[ABCD]$ be a parallelogram with $AB <BC$ and let $E, F$ be points on the circle that passes through $A, B$ and $C$ such that $DE$ and $DF$ are tangents to this circle. Knowing that $\angle ADE = \angle CDF$ , determine $\angle ABC$. [img]https://cdn.artofproblemsolving.com/attachments/5/e/4140b92730e9d382df49ac05ca4e8ba48332dc.png[/img]

2017 Ukrainian Geometry Olympiad, 2

Tags: angle chasing , trapezoid , equal angles , geometry

Point $M$ is the midpoint of the base $BC$ of trapezoid $ABCD$. On base $AD$, point $P$ is selected. Line $PM$ intersects line $DC$ at point $Q$, and the perpendicular from $P$ on the bases intersects line $BQ$ at point $K$. Prove that $\angle QBC = \angle KDA$.

Cono Sur Shortlist - geometry, 2003.G1

Tags: circumcircle , angle chase , equal angles , geometry

Let $O$ be the circumcenter of the isosceles triangle $ABC$ ($AB = AC$). Let $P$ be a point of the segment $AO$ and $Q$ the symmetric of $P$ with respect to the midpoint of $AB$. If $OQ$ cuts $AB$ at $K$ and the circle that passes through $A, K$ and $O$ cuts $AC$ in $L$, show that $\angle ALP = \angle CLO$.

1997 Tournament Of Towns, (554) 4

Tags: angle bisector , equal angles , geometry

Two circles intersect at points $A$ and $B$. A common tangent touches the first circle at point $C$ and the second at point $D$. Let $\angle CBD > \angle CAD$. Let the line $CB$ intersect the second circle again at point $E$. Prove that $AD$ bisects the angle $\angle CAE$. (P Kozhevnikov)

2006 Belarusian National Olympiad, 4

Tags: equal angles , geometry , orthocenter

Given a quadrilateral $ABCD$ with $\angle ABC = \angle ADC$. Let $BM$ be the altitude of the triangle $ABC$, and $M$ belongs to $AC$. Point $M'$ is marked on the diagonal $AC$ so that $$\frac{AM \cdot CM'}{ AM' \cdot CM}= \frac{AB \cdot CD }{ BC \cdot AD}$$ Prove that the intersection point of $DM'$ and $BM$ coincides with the orthocenter of the triangle $ABC$. (M. Zhikhovich)

2017 Czech-Polish-Slovak Match, 2

Tags: concyclic , equal angles , circles , geometry

Let ${\omega}$ be the circumcircle of an acute-angled triangle ${ABC}$. Point ${D}$ lies on the arc ${BC}$ of ${\omega}$ not containing point ${A}$. Point ${E}$ lies in the interior of the triangle ${ABC}$, does not lie on the line ${AD}$, and satisfies ${\angle DBE =\angle ACB}$ and ${\angle DCE = \angle ABC}$. Let ${F}$ be a point on the line ${AD}$ such that lines ${EF}$ and ${BC}$ are parallel, and let ${G}$ be a point on ${\omega}$ different from ${A}$ such that ${AF = FG}$. Prove that points ${D,E, F,G}$ lie on one circle. (Slovakia)

Russian TST 2022, P2

Tags: equal angles , parallelogram , geometry

In parallelogram $ABCD$ with acute angle $A$ a point $N$ is chosen on the segment $AD$, and a point $M$ on the segment $CN$ so that $AB = BM = CM$. Point $K$ is the reflection of $N$ in line $MD$. The line $MK$ meets the segment $AD$ at point $L$. Let $P$ be the common point of the circumcircles of $AMD$ and $CNK$ such that $A$ and $P$ share the same side of the line $MK$. Prove that $\angle CPM = \angle DPL$.

1974 Chisinau City MO, 76

Tags: isosceles , right triangle , equal angles , geometry

Altitude $AH$ and median $AM$ of the triangle $ABC$ satisfy the relation: $\angle ABM = \angle CBH$. Prove that triangle $ABC$ is isosceles or right-angled.

Ukraine Correspondence MO - geometry, 2015.8

Tags: angle bisector , equilateral , geometry , equal angles

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 $EDF$.

2007 Postal Coaching, 5

Tags: concurrency , euler line , equal angles , geometry

Let $P$ be an interior point of triangle $ABC$ such that $\angle BPC = \angle CPA =\angle APB = 120^o$. Prove that the Euler lines of triangles $APB,BPC,CPA$ are concurrent.

2003 Junior Tuymaada Olympiad, 3

Tags: circumcircle , incenter , equal angles , geometry

In the acute triangle $ ABC $, the point $ I $ is the center of the inscribed the circle, the point $ O $ is the center of the circumscribed circle and the point $ I_a $ is the center the excircle tangent to the side $ BC $ and the extensions of the sides $ AB $ and $ AC $. Point $ A'$ is symmetric to vertex $ A $ with respect to the line $ BC $. Prove that $ \angle IOI_a = \angle IA'I_a $.

2020 Ukrainian Geometry Olympiad - April, 4

Tags: geometry , equal angles , angle bisector

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 $\angle AQC$.

2019 Estonia Team Selection Test, 7

Tags: altitude , equal angles , geometry

An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$. The circle with diameter $AC$ intersects the segment $BE$ at point $P$. A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$. Prove that $\angle PQ'Q = \angle PQB$.

1997 Swedish Mathematical Competition, 2

Tags: equal segments , angle , geometry , equal angles

Let $D$ be the point on side $AC$ of a triangle $ABC$ such that $BD$ bisects $\angle B$, and $E$ be the point on side $AB$ such that $3\angle ACE = 2\angle BCE$. Suppose that $BD$ and $CE$ intersect at a point $P$ with $ED = DC = CP$. Determine the angles of the triangle.

2011 Dutch IMO TST, 3

Tags: circles , angle chasing , equal angles , geometry

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 $\angle DPM = \angle BDC$.

2002 Estonia Team Selection Test, 4

Tags: cyclic , angle , equal angles , geometry

Let $ABCD$ be a cyclic quadrilateral such that $\angle ACB = 2\angle CAD$ and $\angle ACD = 2\angle BAC$. Prove that $|CA| = |CB| + |CD|$.

2020 Czech-Austrian-Polish-Slovak Match, 6

Tags: equal angles , fixed , fixed point , geometry

Let $ABC$ be an acute triangle. Let $P$ be a point such that $PB$ and $PC$ are tangent to circumcircle of $ABC$. Let $X$ and $Y$ be variable points on $AB$ and $AC$, respectively, such that $\angle XPY = 2\angle BAC$ and $P$ lies in the interior of triangle $AXY$. Let $Z$ be the reflection of $A$ across $XY$. Prove that the circumcircle of $XYZ$ passes through a fixed point. (Dominik Burek, Poland)

1989 All Soviet Union Mathematical Olympiad, 492

Tags: geometry , equal ratio , similar triangles , equal angles

$ABC$ 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.