MathDB
equal angles wanted and given, square (2016 Kyiv City MO 9.5.1 10.4.1)

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9/5/2020
On the sides ABAB and ADAD of the square ABCDABCD, the points NN and PP are selected, respectively, so that PN=NCPN = NC, the point QQ Is a point on the segment ANAN for which ∠NCB=∠QPN\angle NCB = \angle QPN. Prove that ∠BCQ=12∠PQA\angle BCQ = \tfrac {1} {2} \angle PQA.
geometryequal anglessquare
AB=CD if PS// RQ, angle bisectors in trapezoid (2017 Kyiv City MO 8.4.1)

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9/8/2020
In a trapezoid ABCDABCD with bases ADAD and BCBC, the bisector of the angle ∠DAB\angle DAB intersects the bisectors of the angles ∠ABC\angle ABC and ∠CDA\angle CDA at the points PP and SS, respectively, and the bisector of the angle ∠BCD\angle BCD intersects the bisectors of the angles ∠ABC\angle ABC and ∠CDA\angle CDA at the points QQ and RR, respectively. Prove that if PS∥RQPS\parallel RQ, then AB=CDAB = CD.
geometryangle bisectortrapezoidparallel
collinear wanted, circle intersecting a square (2017 Kyiv City MO 8.4)

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9/8/2020
On the sides BCBC and CDCD of the square ABCDABCD, the points MM and NN are selected in such a way that ∠MAN=45o\angle MAN= 45^o. Using the segment MNMN, as the diameter, we constructed a circle ww, which intersects the segments AMAM and ANAN at points PP and QQ, respectively. Prove that the points B,PB, P and QQ lie on the same line.
geometrysquarecollinear
concyclic wanted, incenter, excenter (2017 Kyiv City MO 9.5)

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9/8/2020
Let II be the center of the inscribed circle of ABCABC and let IAI_A be the center of the exscribed circle touching the side BCBC. Let MM be the midpoint of the side BCBC, and NN be the midpoint of the arc BACBAC of the circumscribed circle of ABCABC . The point TT is symmetric to the point NN wrt point AA. Prove that the points IA,M,I,TI_A,M,I,T lie on the same circle.
(Danilo Hilko)
geometryincenterexcenterConcycliccircumcircle
tangential quadrilateral by medians, isosceles (2017 Kyiv City MO 9.5.1)

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9/8/2020
In the triangle ABCABC, the medians BB1BB_1 and CC1CC_1, which intersect at the point MM, are drawn. Prove that a circle can be inscribed in the quadrilateral AC1MB1AC_1MB_1 if and only if AB=ACAB = AC.
geometrytangential quadrilateralMediansisosceles
AX = AD if <BAX=<CDA,<ABC = <BCD, 2AB = CD (2016 Kyiv City MO 8.5.1)

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9/5/2020
In the quadrilateral ABCDABCD, shown in fig. , the equations are true: ∠ABC=∠BCD\angle ABC = \angle BCD and 2AB=CD2AB = CD. On the side BCBC, a point XX is selected such that ∠BAX=∠CDA\angle BAX = \angle CDA. Prove that AX=ADAX = AD. https://cdn.artofproblemsolving.com/attachments/2/9/0884eb311d1e40300c1e5980fd53eaadfa7a25.png
geometryequal anglesequal segments
angle wanted, midpoints, perpendiculars (2018 Kyiv City MO 7.4.1)

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9/11/2020
In the quadrilateral ABCDABCD point EE - the midpoint of the side ABAB, point FF - the midpoint of the side BCBC, point GG - the midpoint ADAD . It turned out that the segment GEGE is perpendicular to ABAB, and the segment GFGF is perpendicular to the segment BCBC. Find the value of the angle GCDGCD, if it is known that ∠ADC=70∘\angle ADC = 70 {} ^ \circ.
geometryanglesmidpointsperpendicular
angle wanted, circle, midpoint, square related(2018 Kyiv City MO 9.5.1 )

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9/12/2020
Given a circle Γ\Gamma with center at point OO and diameter ABAB. OBDEOBDE is square, FF is the second intersection point of the line ADAD and the circle Γ\Gamma, CC is the midpoint of the segment AFAF. Find the value of the angle OCBOCB.
geometrysquareangles
right angle wanted, <APC=180^o-<ABC,BC = AP, AK=KB+PC (2018 Kyiv City MO 7.4)

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9/11/2020
Inside the triangle ABCABC , the point PP is selected so that BC=APBC = AP and ∠APC=180∘−∠ABC\angle APC = 180 {} ^ \circ - \angle ABC . On the side ABAB there is a point KK , for which AK=KB+PCAK = KB + PC . Prove that ∠AKC=90∘\angle AKC = 90 {} ^ \circ .
(Danilo Hilko)
geometryright angleequal segmentsangles
angle wanted, isosceles, altitudes, rectangle (2018 Kyiv City MO 8.3)

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9/11/2020
In the isosceles triangle ABCABC with the vertex at the point BB, the altitudes BHBH and CLCL are drawn. The point DD is such that BDCHBDCH is a rectangle. Find the value of the angle DLHDLH.
(Bogdan Rublev)
geometryrectangleisoscelesangles
AK bisects BM, <MAC =< PCB, <MPA = <CPK (2018 Kyiv City MO 9.5 10.4.1)

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9/12/2020
Given a triangle ABCABC, the perpendicular bisector of the side ACAC intersects the angle bisector of the triangle AKAK at the point PP, MM - such a point that ∠MAC=∠PCB\angle MAC = \angle PCB, ∠MPA=∠CPK\angle MPA = \angle CPK, and points MM and KK lie on opposite sides of the line ACAC. Prove that the line AKAK bisects the segment BMBM.
(Anton Trygub)
geometrybisects segmentequal angles
<ABM=<MBP wanted, AB //PC , PM_|_BM , median (2021 Kyiv City MO 8.4)

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2/16/2021
Let BMBM be the median of the triangle ABCABC, in which AB>BCAB> BC. Point PP is chosen so that AB∥PCAB \parallel PC and PM⊥BMPM \perp BM. Prove that ∠ABM=∠MBP\angle ABM = \angle MBP.
(Mikhail Standenko)
geometryequal anglesangles
2 equal rectangles inscribed in square (2020 Kyiv City MO 7.4)

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9/19/2020
Given a square ABCDABCD with side 1010. On sides BC and ADAD of this square are selected respectively points EE and FF such that formed a rectangle ABEFABEF. Rectangle KLMNKLMN is located so that its the vertices K,L,MK, L, M and NN lie one on each segments CD,DF,FECD, DF, FE and ECEC, respectively. It turned out that the rectangles ABEFABEF and KLMNKLMN are equal with AB=MNAB = MN. Find the length of segment ALAL.
geometryrectanglesquare
medians AN _|_ CM, when BC=\sqrt2 AC, right triangle (2019 Kyiv City MO 9.2)

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9/16/2020
In a right triangle ABCABC, the lengths of the legs satisfy the condition: BC=2ACBC =\sqrt2 AC. Prove that the medians ANAN and CMCM are perpendicular.
(Hilko Danilo)
geometryMediansperpendicular
concurrency of perp. bisectors (2020 Kyiv City MO 7.4.1)

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9/19/2020
In the quadrilateral ABCDABCD, AB=BCAB = BC . The point EE lies on the line ABAB is such that BD=BEBD= BE and AD⊥DEAD \perp DE. Prove that the perpendicular bisectors to segments AD,CDAD, CD and CECE intersect at one point.
geometryconcurrencyconcurrentperpendicular bisectorequal segments
angle chasing, BK=BO, circumcircle, midpoint, bisector (2020 Kyiv City MO 8.4)

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9/20/2020
Given a triangle ABC,OABC, O is the center of the circumcircle, MM is the midpoint of BC,WBC, W is the second intersection of the bisector of the angle CC with this circle. A line parallel to BCBC passing through WW, intersectsAB AB at the point KK so that BK=BOBK = BO. Find the measure of angle WMBWMB.
(Anton Trygub)
geometrycircumcircleangles
concurrency in cyclic hexagon,AB=BC, CD=DE, EF=FA (2020 Kyiv City MO 8.5.1)

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9/20/2020
Let ABCDEFABCDEF be a hexagon inscribed in a circle in which AB=BC,CD=DEAB = BC, CD = DE and EF=FAEF = FA. Prove that the lines AD,BEAD, BE and CFCF intersect at one point.
geometryconcurrencyconcurrentCyclichexagonequal segments
fixed point for circumcircle, equal angles (2020 Kyiv City MO 9.4)

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9/20/2020
Let the point DD lie on the arc ACAC of the circumcircle of the triangle ABCABC (AB<BCAB < BC), which does not contain the point BB. On the side ACAC are selected an arbitrary point XX and a point X′X' for which ∠ABX=∠CBX′\angle ABX= \angle CBX'. Prove that regardless of the choice of the point XX, the circle circumscribed around △DXX′\vartriangle DXX', passes through a fixed point, which is different from point DD.
(Nikolaev Arseniy)
geometrycircumcirclefixedFixed pointequal angles
angle wanted, cyclic ABCD, AB=BC=CD, AE//CD (2020 Kyiv City MO 9.4.1)

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9/20/2020
The points A,B,C,DA, B, C, D are selected on the circle as followed so that AB=BC=CDAB = BC = CD. Bisectors of ∠ABD\angle ABD and ∠ACD\angle ACD intersect at point EE. Find ∠ABC\angle ABC, if it is known that AE∥CDAE \parallel CD.
geometryanglesequal segments
AM=KB wanted, AK =AO, KM =MC , KM //AC (2021 Kyiv City MO 8.4.1)

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2/16/2021
On the sides ABAB and BCBC of the triangle ABCABC, the points KK and MM are chosen so that KM∥ACKM \parallel AC. The segments AMAM and KCKC intersect at the point OO. It is known that AK=AOAK =AO and KM=MCKM =MC. Prove that AM=KBAM=KB.
geometryequal segments
AB=BQ wanted, AB\\ PC , PM_|_ BM, median (2021 Kyiv City MO 9.5)

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2/15/2021
Let BMBM be the median of the triangle ABCABC, in which AB>BCAB> BC. Point PP is chosen so that AB∥PCAB \parallel PC andPM⊥BM PM \perp BM. The point QQ is chosen on the line BPBP so that ∠AQC=90o\angle AQC = 90^o, and the points BB and QQ lie on opposite sides of the line ACAC. Prove that AB=BQAB = BQ.
(Mikhail Standenko)
geometryequal segmentsmedian
OB_|_CD wanted, concurrent circles (2021 Kyiv City MO 9.5.1)

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2/15/2021
Two circles ω1\omega_1 and ω2\omega_2 intersect at points AA and BB. A line passing through point BB intersects ω1\omega_1 for the second time at point CC and ω2\omega_2 at point DD. The line ACAC intersects circle ω2\omega_2 for the second time at point FF, and the line ADAD intersects the circle ω1\omega_1 for the second time at point EE . Let point OO be the center of the circle circumscribed around △AEF\vartriangle AEF. Prove that OB⊥CDOB \perp CD.
geometrycirclesperpendicular
fixed point, fixed length, fixed angle (2010 Kyiv City MO Round2 10.4 11.4)

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8/2/2020
The points A≠BA \ne B are given on the plane. The point CC moves along the plane in such a way that ∠ACB=α\angle ACB = \alpha , where α\alpha is the fixed angle from the interval (0o,180o0^o, 180^o). The circle inscribed in triangle ABCABC has center the point II and touches the sides AB,BC,CAAB, BC, CA at points D,E,FD, E, F accordingly. Rays AIAI and BIBI intersect the line EFEF at points MM and NN, respectively. Show that: a) the segment MNMN has a constant length, b) all circles circumscribed around triangle DMNDMN have a common point
fixedgeometryFixed pointlengthanglescircumcircle
perpendicular wanted circumcircle related (2018 Kyiv City MO Round2 10.3.1)

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9/14/2020
The point OO is the center of the circumcircle of the acute triangle ABCABC. The line ACAC intersects the circumscribed circle ΔABO\Delta ABO for second time at the point XX. Prove that XO⊥BCXO \bot BC.
geometrycircumcircleperpendicular
concurrency, // diameters of 3 tangent circles (2011 Kyiv City MO Round2 11.4)

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7/31/2020
Let three circles be externally tangent in pairs, with parallel diameters A1A2,B1B2,C1C2A_1A_2, B_1B_2, C_1C_2 (i.e. each of the quadrilaterals A1B1B2A2A_1B_1B_2A_2 and A1C1C2A2A_1C_1C_2A_2 is a parallelogram or trapezoid, which segment A1A2A_1A_2 is the base). Prove that A1B2,B1C2,C1A2A_1B_2, B_1C_2, C_1A_2 intersect at one point.
(Yuri Biletsky )
geometryconcurrencyconcurrenttangent circles