geometry problems from Euler Olympiads (remote (1st), regional (2nd) and final (3rd) rounds , from Russia) with aops links in the names
Given a convex quadrilateral ABSC. A point P is chosen on the diagonal BC so that AP = CP > BP. Point Q is symmetric to the point P with respect to the center of the diagonal BC , R is symmetric the point Q with respect to the line AC. It turned out that \angle SAB = \angle QAC and \angle SBC = \angle BAC. Prove that SA = SR.
source: matol.kz/nodes/74 (official site)
Final Round (3rd)
2009 Euler Olympiad P3
In the triangle ABC , the sides AB and BC are equal. The point D inside the triangle is such that the angle ADC is twice as large as the angle ABC . Prove that the double distance from the point B to the line bisecting the angles externally to the angle ADC is AD + DC .
In the triangle ABC , the sides AB and BC are equal. The point D inside the triangle is such that the angle ADC is twice as large as the angle ABC . Prove that the double distance from the point B to the line bisecting the angles externally to the angle ADC is AD + DC .
2009 Euler Olympiad P6
In the convex quadrilateral ABCD the relations AB = BD are satisfied; \angle ABD = \angle DBC . On the diagonal BD there was a point K such that BK = BC . Prove that \angle KAD = \angle KCD .
2010 Euler Olympiad P3
In the quadrilateral ABCD , the side AB is equal to the diagonal AC and is perpendicular to the side AD , and the diagonal AC is perpendicular to the side CD . On the side AD , a point K is taken such that AC = AK . The bisector of the angle ADC intersects BK at the point M . Find the angle ACM .
In the convex quadrilateral ABCD the relations AB = BD are satisfied; \angle ABD = \angle DBC . On the diagonal BD there was a point K such that BK = BC . Prove that \angle KAD = \angle KCD .
2010 Euler Olympiad P3
In the quadrilateral ABCD , the side AB is equal to the diagonal AC and is perpendicular to the side AD , and the diagonal AC is perpendicular to the side CD . On the side AD , a point K is taken such that AC = AK . The bisector of the angle ADC intersects BK at the point M . Find the angle ACM .
2010 Euler Olympiad P6
In the convex quadrilateral ABCD , the angles B and D are equal, CD = 4BC , and the bisector of the angle A passes through the middle of the side CD . What can the AD / AB ratio be?
2011 Euler Olympiad P4
Inside the convex quadrilateral ABCD , in which AB = CD , the point P is chosen in such a way that the sum of the angles PBA and PCD is 180 degrees. Prove that PB+PC<AD.
In the convex quadrilateral ABCD , the angles B and D are equal, CD = 4BC , and the bisector of the angle A passes through the middle of the side CD . What can the AD / AB ratio be?
2011 Euler Olympiad P4
Inside the convex quadrilateral ABCD , in which AB = CD , the point P is chosen in such a way that the sum of the angles PBA and PCD is 180 degrees. Prove that PB+PC<AD.
2011 Euler Olympiad P6
The convex pentagon ABCDE is such that AB \parallel CD , BC \parallel AD , AC \parallel DE , CE \perp BC . Prove that EC is the bisector of the angle BED.
2012 Euler Olympiad P1
On the side BC of the triangle ABC the point D is taken in such a way that the perpendicular bisector of segment AD passes through the center of the circle inscribed in the triangle ABC . Prove that this perpendicular passes through a vertice of the triangle ABC .
2019 Euler Olympiad P4The convex pentagon ABCDE is such that AB \parallel CD , BC \parallel AD , AC \parallel DE , CE \perp BC . Prove that EC is the bisector of the angle BED.
2012 Euler Olympiad P1
On the side BC of the triangle ABC the point D is taken in such a way that the perpendicular bisector of segment AD passes through the center of the circle inscribed in the triangle ABC . Prove that this perpendicular passes through a vertice of the triangle ABC .
2012 Euler Olympiad P7
The angles of the triangle ABC satisfy the condition 2 \angle A + \angle B = \angle C. Inside this triangle, the point K is chosen on the bisector of the angle A such that BK = BC. Prove that \angle KBC = 2 \angle KBA.
2013 Euler Olympiad P3
The diagonals of the convex quadrilateral ABCD are equal and intersect at the point O . The point P inside the triangle AOD is such that CD \parallel BP and AB \parallel CP . Prove that the point P lies on the bisector of the angle AOD .
The angles of the triangle ABC satisfy the condition 2 \angle A + \angle B = \angle C. Inside this triangle, the point K is chosen on the bisector of the angle A such that BK = BC. Prove that \angle KBC = 2 \angle KBA.
2013 Euler Olympiad P3
The diagonals of the convex quadrilateral ABCD are equal and intersect at the point O . The point P inside the triangle AOD is such that CD \parallel BP and AB \parallel CP . Prove that the point P lies on the bisector of the angle AOD .
2013 Euler Olympiad P6
In the convex quadrilateral ABCD in which AB = CD , the points K and M are chosen on the sides AB and CD , respectively. It turned out that AM = KC , BM = KD . Prove that the angle between the lines AB and KM is equal to the angle between the lines KM and CD .
In the convex quadrilateral ABCD in which AB = CD , the points K and M are chosen on the sides AB and CD , respectively. It turned out that AM = KC , BM = KD . Prove that the angle between the lines AB and KM is equal to the angle between the lines KM and CD .
On the side AB of a triangle ABC with an angle of 100 ^\circ at the vertice C, points P and Q such that AP = BC and BQ = AC . Let M, N, K be midpoints of AB, CP, CQ, respectively. Find the angle NMK.
A diagonal of a convex 101-gon will be called [i]principal [/i] if there are 50 vertices on one side of it and 49 vertices on the other. Several principal diagonals have been chosen that have no common ends. Prove that the sum of the lengths of these diagonals is less than the sum of the lengths of the remaining principal diagonals.
In the triangle ABC , side AB is greater than side BC. On the extension of the side BC beyond the point C, noted the point N so that 2BN = AB+BC. Let BS be the angle bisector of ABC, M be the midpoint of the side AC, а L be a point on the side BS such that ML \parallel AB . Prove that 2LN = AC.
2015 Euler Olympiad P8
Let CK be angle bisector of the triangle ABC. On the sides BC and AC, are chosen points L and T respectively, such that CT = BL and TL = BK. Prove that the triangle LTC is similar to the original one.
2015 Euler Olympiad P8
Let CK be angle bisector of the triangle ABC. On the sides BC and AC, are chosen points L and T respectively, such that CT = BL and TL = BK. Prove that the triangle LTC is similar to the original one.
An equilateral triangle ABC. is given. Point D is chosen on the extension of the side AB beyond the point A, point E on the extension of BC beyond the point C, and point F on the extension of AC beyond the point C so that CF = AD and AC+EF = DE. Find the angle BDE.
2016 Euler Olympiad P8
A parallelogram ABCD. is given. On the sides AB and BC and the extension of the side CD beyond the point D , the points K, L and M are chosen, and so that the triangles KLM and BCA are congruent (precisely with such a correspondence of the vertices). The segment KM intersects the segment AD at the point N. Prove that LN \parallel AB.
2016 Euler Olympiad P8
A parallelogram ABCD. is given. On the sides AB and BC and the extension of the side CD beyond the point D , the points K, L and M are chosen, and so that the triangles KLM and BCA are congruent (precisely with such a correspondence of the vertices). The segment KM intersects the segment AD at the point N. Prove that LN \parallel AB.
Diagonals of a convex quadrilateral ABCD intersect at a point E. It is known that AB=BC=CD=DE=1. Prove that AD<2
2017 Euler Olympiad P6
2017 Euler Olympiad P6
In a convex quadrilateral ABCD angles A and C are equal 100^\circ. On the sides AB and BC points X and Y are selected so that AX = CY . It turned out that the line YD is parallel to the angle bisector of the angle ABC. Find the angle AXY.
2018 Euler Olympiad P3
Diagonals of a convex quadrilateral ABCD are equal and intersect at a point K . Inside the triangles AKD and BKC select points P and Q so that \angle KAP = \angle KDP = \angle KBQ = \angle KCQ. Prove that line PQ is parallel to the bisector of the angle AKD.
Diagonals of a convex quadrilateral ABCD are equal and intersect at a point K . Inside the triangles AKD and BKC select points P and Q so that \angle KAP = \angle KDP = \angle KBQ = \angle KCQ. Prove that line PQ is parallel to the bisector of the angle AKD.
2018 Euler Olympiad P8
Vertex F of a parallelogram ACEF lies on the side BC of a parallelogram ABCD . It is known that AC = AD and AE = 2CD. Prove that \angle CDE = \angle BEF.
Vertex F of a parallelogram ACEF lies on the side BC of a parallelogram ABCD . It is known that AC = AD and AE = 2CD. Prove that \angle CDE = \angle BEF.
Given a convex quadrilateral ABSC. A point P is chosen on the diagonal BC so that AP = CP > BP. Point Q is symmetric to the point P with respect to the center of the diagonal BC , R is symmetric the point Q with respect to the line AC. It turned out that \angle SAB = \angle QAC and \angle SBC = \angle BAC. Prove that SA = SR.
2019 Euler Olympiad P6
The points M and N are the midpoints of the sides AB and BC respectively of the triangle ABC. The point D is marked on the extension of the segment CM for the point M . It turned out that BC = BD = 2 and AN = 3. Prove that \angle ADC = 90^\circ.
2020 Euler Olympiad P3
Given is a triangle ABC with 2<B=180°+<A. Point K is in the interior of the triangle, so that <ACK=2<BCK and L is on AB, so that BK=KL. Prove that CK+AL=AC
The points M and N are the midpoints of the sides AB and BC respectively of the triangle ABC. The point D is marked on the extension of the segment CM for the point M . It turned out that BC = BD = 2 and AN = 3. Prove that \angle ADC = 90^\circ.
2020 Euler Olympiad P3
Given is a triangle ABC with 2<B=180°+<A. Point K is in the interior of the triangle, so that <ACK=2<BCK and L is on AB, so that BK=KL. Prove that CK+AL=AC
On every side of convex 100-gon are chosen two points which divide every side into 3 equal parts. We delete all vertices of the 100-gon. Prove that by knowing where the newly added points are, we can restore the 100-gon.
Points P and Q are selected on sides AB and BC of triangle ABC respectively. The segments CP and AQ intersect at point R. It turned out that AR = CR = PR + QR. Prove that from the segments AP, CQ and PQ you can make a triangle, one of the angles of which is equal to angle B.
The diagonals of the trapezoid ABCD (AD\parallel BC) meet at point K. Inside the triangle ABK, there is a point M such that \angle MBC = \angle MAD, \angle MCB = \angle MDA. Prove that line MK is parallel to the bases of the trapezoid.
Point D is marked on side BC of triangle ABC. A point P is chosen on the side AB. The segments PC and AD intersect at the point Q. Point R is the midpoint of segment AP. Prove that there exists a fixed point X through which the line RQ passes for any choice of the point P.
In a convex pentagon ABCDE, the diagonals AD and CE intersect at the point X. It turned out that ABCX is a parallelogram and BD = CX, BE = AX. Prove that AE = CD.
Regional Round (2nd)
Point K is the midpoint of hypotenuse AB of right-angled isosceles triangle ABC . Points L and M are chosen on legs BC and AC , respectively, such that BL = CM . Prove that LMK is also a right-angled isosceles triangle.
In the convex quadrilateral ABCD , a point of the diagonal AC lies on the perpendicular bisectors of the sides AB and CD , and a point of the diagonal BD lies on the perpendicular bisectors of the sides AD and BC . Prove that ABCD is a rectangle.
On the hypotenuse BC of the right-angled triangle ABC , the point K is chosen so that AB = AK . The segment AK intersects the angle bisector CL in its midpoint . Find the acute angles of the triangle ABC .
The bisectors of the angles A and C of the trapezoid ABCD intersect at the point P , and the bisectors of the angles B and D intersect at the point Q , different from P . Prove that if the segment PQ is parallel to the base AD , then the trapezoid is isosceles.
You are given a convex quadrilateral ABCD such that AD = AB + CD . It turned out that the bisector of the angle A passes through the midpoiint of the side BC . Prove that the bisector of D also passes through the midpoint of BC .
In triangle ABC the points M and N are the midpoints of the sides AC and AB , respectively. A point P is chosen on the median BM that does not lie on CN . It turned out that PC = 2PN . Prove that AP = BC .
The trapezoid ABCD with bases AD and BC is such that the angle ABD is right and BC + CD = AD . Find the ratio of the bases AD: BC .
In the convex quadrilateral ABCD , the angles ABC and ADC are right. Points K , L , M , N are taken on the sides AB , BC , CD , DA , respectively, so that KLMN is a rectangle. Prove that the midpoint of the diagonal AC is equidistant from the lines KL and MN .
The point M is marked on the segment AB . The points P and Q are the midpoints of the segments AM and BM respectively, the point O is the midpoint of the segment PQ . Let's choose point C so that the angle ACB is right. Let MD and ME be the perpendiculars dropped from the point M onto the lines CA and CB , and F the midpoint of the segment DE . Prove that the length of the segment OF does not depend on the choice of the point C .
Given an acute-angled triangle ABC . The altitude AA_1 is extended beyond the top A to the segment AA_2 = BC . The altitude CC_1 is extended beyond the top C to the segment CC_2 = AB . Find the angles of the triangle A_2BC_2 .
On the side AC of the triangle ABC , a point D is chosen such that BD = AC . The median AM of this triangle intersects the segment BD at the point K . It turned out that DK = DC . Prove that AM + KM = AB .
Given a convex pentagon ABCDE , where the line BE is parallel to the line CD and the segment BE is shorter than the segment CD . The points F and G are selected inside the pentagon in such a way that ABCF and AGDE are parallelograms. Prove that CD = BE + FG .
The perpendicular bisectors to the sides AB and BC of the convex quadrilateral ABCD meet the sides CD and DA at the points P and Q , respectively. It turned out that \angle APB = \angle BQC . Inside the quadrangle, a point X is selected such that QX \parallel AB and PX \parallel BC . Prove that the line BX bisects the diagonal of AC ..
In trapezoid ABCD , where AD \parallel BC , the angle B is equal to the sum of the angles A and D . On the extension of the segment CD beyond the vertex D , draw the segment DK = BC . Prove that AK = BK .
In trapezoid ABCD , point M is the midpoint of the base of AD . It is known that \angle ABD = 90 ^\circ and BC = CD . On the segment BD , a point F is chosen such that \angle BCF = 90 ^\circ . Prove that MF \perp CD .
The points M and N are the midpoints of the angle bisectors AK and CL of the triangle ABC , respectively. Prove that an angle ABC is right if and only if \angle MBN = 45 ^\circ .
On the lateral sides AB and AC of an isosceles triangle ABC , the points P and Q are selected, respectively, so that PQ \parallel BC . On the angle bisectors of triangles ABC and APQ outgoing from the vertices B and Q , the points X and Y , respectively, are chosen so that XY \parallel BC . Prove that PX = CY .
In convex quadrilateral ABCD , the bisector of angle B passes through the midpoint of side AD , and \angle C = \angle A + \angle D . Find the angle ACD .
Inside the parallelogram ABCD , the point E is chosen so that AE = DE and \angle ABE = 90 ^ \circ. Point M is the midpoint of the segment BC. Find the angle DME.
Point D is selected on the angle bisector AL of the triangle ABC . It is known that \angle BAC = 2 \alpha, \angle ADC = 3 \alpha, \angle ACB = 4 \alpha. Prove, that BC + CD = AB.
The perimeter of the triangle ABC is 2. The point P is marked on the side AC , and the point Q is marked on the segment CP so that 2AP = AB and 2QC = BC. Prove that the perimeter of the triangle BPQ is greater than 1.
Point N is the midpoint of side BC of triangle ABC, in which \angle ACB = 60 ^\circ . Point M on side AC is such that AM = BN. Point K is the midpoint of segment BM. Prove that AK = KC. ympi
The bisector of the angle A of the convex quadrilateral ABCD meets the side CD at the point K. It turns out that DK = BC and KC + AB = AD. Prove that \angle BCD = \angle ADC.
On the midline of an equilateral triangle ABC, parallel to the side BC, point D is taken. Point E on the extension of side BA beyond point A is such that \angle ECA = \angle DCA. Point F on the extension of the side CA beyond the point A is such that \angle FBA = \angle DBA. Prove that the point A lies on the midline of the triangle DEF, parallel to the side EF.
CL is the angle bisector of triangle ABC. CLBK is a parallelogram. Line AK intersects
segment CL at point P. It turned out that point P is equidistant from the diagonals of the
parallelogram CLBK. Prove that AK \ge CL.
2021 Euler Olympiad Regional 2.2
Point M is the midpoint of side AC of equilateral triangle ABC. Points P and R on the
segments AM and BC are respectively chosen so that AP = BR. Find the sum of the angles
ARM, PBM, and BMR.
Angle bisectors BK and CL are drawn in triangle ABC. A point N is marked on the segment
BK so that LN \parallel AC. It turned out that NK = LN. Find the measure of the angle ABC.
2022 Euler Olympiad Regional 1.5
What is the largest n for which there exists a convex n-gon whose diagonal lengths are at most two
different values?
2022 Euler Olympiad Regional 2.2
Is there a triangle in which the lengths of non-coinciding medians and altitudes drawn from one of its
vertices are respectively equal to the lengths of the two sides of this triangle?
Given an isosceles triangle ABC (AC = BC) . Points A_1 , B_1 and C_1 are marked on the sides BC , AC , AB , respectively. It turned out that C_1B_1 is perpendicular to AC , B_1A_1 is perpendicular to BC and B_1A_1 = B_1C_1 . Prove that A_1C_1 is perpendicular to AB .
The two angle bisectors of the triangle intersect at an angle of 60 degrees. Prove that one of the angles of this triangle is 60 degrees.
The length of the rectangle has been decreased by 10 \% and the width has been decreased by 20 \% . In this case, the perimeter of the rectangle has decreased by 12 \% . By what percentage will the perimeter of the rectangle decrease if its length is reduced by 20 \% , and reduce the width by 10 \% ?
Can the distances from a point on the plane to the vertices of a certain square be equal to 1, 1, 2 and 3?
The point D lies on the hypotenuse AB of the right-angled triangle ABC , but does not coincide with its middle. Prove that there are no equal segments among the segments AD , BD and CD .
In triangle ABC , the median BM is half the side of AB and makes an angle of 40 degrees with it. Find the angle ABC .
In a right-angled triangle, the altitude dropped on the hypotenuse is four times shorter than the hypotenuse. Find the acute angles of the triangle.
There are four properties of quads:
(1) opposite sides are equal in pairs;
(2) two opposite sides are parallel;
(3) some two adjacent sides are equal;
(4) the diagonals are perpendicular and divided by the point of intersection in the same ratio.
One of these two quadrilaterals has some two of these properties, the other two others.
Prove that one of these two quads is a rhombus.
The points E and F are the midpoints of the sides BC and CD respectively of the rectangle ABCD . Prove that AE <2EF .
Inside the angle AOB , equal to 120 ^\circ , the rays OC and OD are drawn so that each of them is the bisector of one of the angles obtained in the drawing. Find the value of angle AOC . Indicate all possible options.
An arbitrary point D on the median BM is marked in the triangle ABC . Then a line parallel to AB was drawn through D , and a line parallel to BM through C . These lines intersect at the point E . Prove that BE = AD .
The point E is marked on the side BC of the triangle ABC , and the point F is marked on the angle bisector BD in such a way that EF \parallel AC and AF = AD . Prove that AB = BE .
Divide a 30 ^ \circ right-angled triangle into two smaller triangles so that some median of one of these triangles is parallel to one of the angle bisectors of the second triangle.
Given a pentagon ABCDE such that AB = BC = CD = DE , \angle B = 96 ^ \circ \angle C = \angle D = 108 ^ \circ . Find \angle E .
In triangle ABC , the angle C is three times the angle A , and the side AB is twice the side BC . Prove that the angle ABC is 60 degrees.
The angle bisector BL is drawn in the triangle ABC , and the point K is chosen on its extension beyond the point L , for which LK = AB . It turned out that AK \parallel BC . Prove that AB> BC .
Inside the acute angle BAC , we took such a point D that the angle CAD is twice the angle BAD . Could point D be twice as far from line AC as from line AB ?
Can five identical rectangles with a perimeter of 10 make one rectangle with a perimeter of 22?
In a triangle ABC , AC = 1 , AB = 2 , O is the intersection point of the angle bisectors. A segment passing through point O parallel to side BC intersects sides AC and AB at points K and M , respectively. Find the perimeter of the triangle AKM .
In a triangle ABC AB = BC . Points D , E , and F are marked on the rays CA , AB and BC , respectively, so that AD = AC , BE = BA , CF = CB . Find the sum of the angles ADB , BEC and CFA .
Point E is the midpoint of the base AD of trapezoid ABCD . The segments BD and CE meet at the point F . It is known that AF \perp BD . Prove that BC = FC .
Point D lies inside triangle ABC . Could it be that the shortest side of triangle BCD is 1, the shortest side of triangle ACD is 2, and the shortest side of triangle ABD is 3?
A smaller square was cut out of the square, one of the sides of which lies on the side of the original square. The perimeter of the resulting octagon is 40 \% larger than the perimeter of the original square. How many percent is its area less than the original square?
In parallelogram ABCD with side AB = 1 , point M is the midpoint of side BC , and the angle AMD is 90 degrees. Find the side BC .
The diagonals AD and BE of the convex pentagon ABCDE meet at the point P . It is known that AC = CE = AE , \angle APB = \angle ACE and AB + BC = CD + DE . Prove that AD = BE .
The square is cut into rectangles of equal area as shown in the figure. Find the area of the square if the segment AB is 1.
Inside the angle BAC , equal to 45 ^\circ , point D is taken so that each of the angles ADB and ADC is 45 ^\circ . Points D_1 and D_2 are symmetric to point D with respect to lines AB and AC , respectively. Prove that the points D_1 , D_2 , B and C are collinear.
Points D and E are marked on the side AC of triangle ABC with an angle of 120 degrees at the vertex B such that AD = AB and CE = CB . From the point D , the perpendicular DF is dropped to the line BE . Find the ratio BD / DF .
On the sides AB , BC , CD and DA of the quadrilateral ABCD , the points K , L , M , N are selected, respectively, so that AK = AN , BK = BL , CL = CM , DM = DN and KLMN is a rectangle. Prove that ABCD is a rhombus.
The median BM is drawn in the triangle ABC . It is known that \angle ABM = 40 ^ \circ and \angle CBM = 70 ^ \circ . Find the ratio AB: BM .
In the triangle ABC the angle bisector BD was drawn, in the triangle BDC the angle bisector DE , and in the triangle DEC the angle bisector EF . It turned out that the lines BD and EF are parallel. Prove that the angle ABC is twice the angle BAC .
The bisectors of the angles A and C cut the non-isosceles triangle ABC into a quadrilateral and three triangles, and among these three triangles there are two isosceles. Find the angles of the triangle ABC .
The squares with sides 11, 9, 7, and 5 are located approximately as in the picture below. It turned out that the area of the gray parts is twice as large as the area of the black parts. Find the area of the white parts.
In triangle ABC , point M is the midpoint of AC , in addition, BC = 2AC / 3 and \angle BMC = 2 \angle ABM . Find the ratio AM / AB .
Can the median and the angle bisector drawn from the vertex A of an acute-angled triangle ABC divide the altitude BH of this triangle into three equal parts?
In triangle ABC , the angle C is 2 times the angle B , CD is the angle bisector. From the midpoint M of the side BC the perpendicular MH is drawn on the segment CD . There is a point K on the side AB such that KMH is an equilateral triangle. Prove that the points M , H and A are collinear.
The diagonals of the parallelogram ABCD meet at the point O . The point P such that DOCP is also a parallelogram ( CD is its diagonal). Let Q denote the intersection point of BP and AC , and let R be the intersection point of DQ and CP . Prove that PC = CR .
In trapezoid ABCD (AD \parallel BC) AD = 2 , BC = 1 , \angle ABD = 90 ^ \circ . Find the side CD .
The teacher drew a rectangle ABCD on the blackboard. Student Petya divided this rectangle into two rectangles with a straight line parallel to the side AB . It turned out that the areas of these parts are 1: 2, and the perimeters are 3: 5 (in the same order). Pupil Vasya divided this rectangle into two parts by a straight line parallel to the BC side. The area of the new parts is also 1: 2. What is the ratio of their perimeters ?
Point K is taken in triangle ABC on side BC . KM and KP are the angle bisectors of triangles AKB and AKC , respectively. It turned out that the diagonal MK divides the quadrilateral BMPK into two equal triangles. Prove that M is the midpoint of AB .
Petya marks four points on the plane so that all of them cannot be crossed out by two parallel straight lines. Vasya chooses two of the lines passing through pairs of points, measures the angle between them and pays Petya an amount equal to the degree measure of the angle. What is the largest amount Petya can guarantee himself?
The angle bisector BD was drawn in the triangle ABC , and in the triangles ABD and CBD the angle bisectors DE and DF , respectively. It turned out that EF \parallel AC . Find the angle DEF .
Borya drew nine segments, three of which are equal to three altitudes of the ABC triangle, three to three angle bisectors, three to three medians. It turned out that for any of the drawn segments, among the other eight, there is an equal to it. Prove that triangle ABC is isosceles.
The stick is broken into 15 parts so that no three parts can be used to form a triangle. Prove that among the pieces there is one that is longer than a third of the original stick.
In a convex pentagon ABCDE , AB is parallel to DE , CD = DE , CE is perpendicular to BC and AD . Prove that the line passing through A parallel to CD , the line passing through B parallel to CE , and the line passing through E parallel to BC , intersect at one point.
In trapezoid ABCD , the side of AB is equal to the diagonal of BD. The point M is the midpoint of the diagonal AC . Line BM intersects CD at point E. Prove that BE = CE.
The point P is located inside the triangle ABC . On the side BC , a point H is selected that does not coincide with the midpoint of the side. It turned out that the bisector of the angle AHP is perpendicular to the side BC , the angle ABC is equal to the angle HCP and BP = AC . Prove that BH = AH .
In trapezoid ABCD , the base AD is greater than the lateral side CD. The bisector of the angle D meets the side AB at the point K. Prove that AK> KB.
Inside the trapezoid ABCD (BC \parallel AD), where AD = 2BC, is the point F, for which AB = FB. The point M is the midpoint of the segment FD. Prove that CM \perp FA.
The points D and E lie on the extensions of the sides AB and BC of an acute-angled triangle ABC beyond the points B and C , respectively. The points M and N are the midpoints of the segments AE and DC. Prove that MN> AD / 2.
In an acute-angled triangle ABC , the angle at the vertex A is 45 degrees. Prove that the perimeter of this triangle is less than twice the sum of its altitudes drawn from the vertices B and C .
In triangle ABC (\angle C = 90^o) on leg BC, points K and L are marked such that \angle CAK = \angle KAL = \angle LAB. A point M is marked on the hypotenuse AB such that ML = KL. Prove that the perpendicular from the point C to the line AK does not bisect the segment ML.
AH is the altitude of an isosceles triangle ABC (AB = BC). HK is the altitude of the triangle AHB. It turned out that 4HK = AB. What could be the measure of the angle ABC?
In trapezoid ABCD, the bisector of angle B intersects the base AD at point L. Point M is the midpoint of side CD. A straight line parallel to BM and passing through L intersects side AB at point K. It turned out that the angle BLM is a straight line. Find the ratio BK / KA .
Given a triangle ABC in which AB = BC. On the side BC there is a point D such that CD = AC. The point E on the ray DA is such that DE = AC. Which segment is longer , EC or AC?
ABCD is a convex quadrilateral, where AB = 7, BC = 4, AD = DC, \angle ABD = \angle DBC. The point E on the segment AB is such that \angle DEB = 90^\circ. Find the length of the segment AE.
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