Olympiad problems

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 Saudi Arabia JBMO TST, 2

Tags: geometry , equal segments , equal angles

In a triangle $ABC$, let $K$ be a point on the median $BM$ such that $CM = CK$. It turned out that $\angle CBM = 2\angle ABM$. Show that $BC = KM$.

Swiss NMO - geometry, 2019.7

Tags: geometry , equal angles , equal segments , angle

Let $ABC$ be a triangle with $\angle CAB = 2 \angle ABC$. Assume that a point $D$ is inside the triangle $ABC$ exists such that $AD = BD$ and $CD = AC$. Show that $\angle ACB = 3 \angle DCB$.

2015 Caucasus Mathematical Olympiad, 5

Tags: midpoint , altitude , geometry , equal angles , equal segments

Let $AA_1$ and $CC_1$ be the altitudes of the acute-angled triangle $ABC$. Let $K,L$ and $M$ be the midpoints of the sides $AB,BC$ and $CA$ respectively. Prove that if $\angle C_1MA_1 =\angle ABC$, then $C_1 K = A_1L$.

2019 IFYM, Sozopol, 4

Tags: geometry , equal angles , power of a point

The diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$ intersect in point $M$. The angle bisector of $\angle ACD$ intersects the ray $\overrightarrow{BA}$ in point $K$. If $MA.MC+MA.CD=MB.MD$, prove that $\angle BKC=\angle CDB$.

1998 Tournament Of Towns, 2

Tags: geometry , parallelogram , equal segments , equal angles

$ABCD$ is a parallelogram. A point $M$ is found on the side $AB$ or its extension such that $\angle MAD = \angle AMO$ where $O$ is the intersection point of the diagonals of the parallelogram. Prove that $MD = MG$. (M Smurov)

1997 Swedish Mathematical Competition, 2

Tags: geometry , equal angles , equal segments , angle

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.

2018 Irish Math Olympiad, 8

Tags: equal angles , geometry , equilateral triangle

Let $M$ be the midpoint of side $BC$ of an equilateral triangle $ABC$. The point $D$ is on $CA$ extended such that $A$ is between $D$ and $C$. The point $E$ is on $AB$ extended such that $B$ is between $A$ and $E$, and $|MD| = |ME|$. The point $F$ is the intersection of $MD$ and $AB$. Prove that $\angle BFM = \angle BME$.

2021 Bosnia and Herzegovina Junior BMO TST, 3

Tags: geometry , right angle , equal angles

In the convex quadrilateral $ABCD$, $AD = BD$ and $\angle ACD = 3 \angle BAC$. Let $M$ be the midpoint of side $AD$. If the lines $CM$ and $AB$ are parallel, prove that the angle $\angle ACB$ is right.

1999 Portugal MO, 6

Tags: equal angles , angle , geometry

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 $\angle BAC$ .

2019 Romania National Olympiad, 3

Tags: geometry , 3d geometry , prism , equal angles , angle

In the regular hexagonal prism $ABCDEFA_1B_1C_1D_1E_1F_1$, We construct $, Q$, the projections of point $A$ on the lines $A_1B$ and $A_1C$ repsectilvely. We construct $R,S$, the projections of point $D_1$ on the lines $A_1D$ and $C_1D$ respectively. a) Determine the measure of the angle between the planes $(AQP)$ and $(D_1RS)$. b) Show that $\angle AQP = \angle D_1RS$.