Found problems: 509
Three different points $A,B$ and $C$ lie on a circle with center $M$ so that $| AB | = | BC |$. Point $D$ is inside the circle in such a way that $\vartriangle BCD$ is equilateral. Let $F$ be the second intersection of $AD$ with the circle . Prove that $| F D | = | FM |$.
Let $ABC$ be a triangle where $|AB| = |AC|$. Points $P$ and $Q$ are different from the vertices of the triangle and lie on the sides $AB$ and $AC$, respectively. Prove that the circumcircle of the triangle $APQ$ passes through the circumcenter of $ABC$ if and only if $|AP| = |CQ|$.
Given an angle with the vertex $O$ and a circle touching its sides in the points $A$ and $B$. A ray is drawn from the point $A$ parallel to $[OB)$. It intersects with the circumference in the point $C$. The segment $[OC]$ intersects the circumference in the point $E$. The straight lines $(AE)$ and $(OB)$ intersect in the point $K$. Prove that $|OK| = |KB|$.
Consider the triangle $ABC$, where $\angle B= 30^o, \angle C = 15^o$, and $M$ is the midpoint of the side $[BC]$. Let point $N \in (BC)$ be such that $[NC] = [AB]$. Show that $[AN$ is the angle bisector of $MAC$
Let $ABCD$ be a cyclic quadrilateral with $\angle ABC = 60^o$ and $| BC | = | CD |$. Prove that $|CD| + |DA| = |AB|$
In a triangle $ABC$ let $E$ be the midpoint of the side $AC$ and $F$ the midpoint of the side $BC$. Let $G$ be the foot of the perpendicular from $C$ to $ AB$. Show that $\vartriangle EFG$ is isosceles if and only if $\vartriangle ABC$ is isosceles.
In a right triangle $ABC$, $K$ is the midpoint of the hypotenuse $AB$ and $M$ such a point on the $BC$ that $| B M | = 2 | MC |$. Prove that $\angle MAB = \angle MKC$.
Given a triangle $ABC$ with $AB = AC$, angle $\angle A = 100^o$ and $BD$ bisector of angle $\angle B$. Prove that $$BC = BD + DA.$$
In the $9$-gon $ABCDEFGHI$, all sides have equal lengths and all angles are equal. Prove that $|AB| + |AC| = |AE|$.
[img]https://cdn.artofproblemsolving.com/attachments/6/2/8c82e8a87bf8a557baaf6ac72b3d18d2ba3965.png[/img]
Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles.
[asy]
unitsize(1.5 cm);
pair A, B, C, D, G, H;
A = (0,0);
B = (2,0);
D = (0.5,1.5);
C = B + D - A;
G = reflect(A,B)*(C) + C - B;
H = reflect(B,C)*(H) + A - B;
draw(H--A--D--C--G);
draw(interp(A,G,-0.1)--interp(A,G,1.1));
draw(interp(C,H,-0.1)--interp(C,H,1.1));
draw(D--G--H--cycle, dashed);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, E);
dot("$D$", D, NW);
dot("$G$", G, NE);
dot("$H$", H, SE);
[/asy]
There are 2022 marked points on a straight line so that every two adjacent points are the same distance apart. Half of all the points are coloured red and the other half are coloured blue. Can the sum of the lengths of all the segments with a red left endpoint and a blue right endpoint be equal to the sum of the lengths of all the segments with a blue left endpoint and a red right endpoint?
On the legs $AC, BC$ of a right triangle $\vartriangle ABC$ select points $M$ and $N$, respectively, so that $\angle MBC = \angle NAC$. The perpendiculars from points $M$ and $C$ on the line $AN$ intersect $AB$ at points $K$ and $L$, respectively. Prove that $KL=LB$.
(O. Clurman)
The centre of circle $1$ lies on circle $2$. $A$ and $B$ are the intersection points of the circles. The tangent line to circle $2$ at point $B$ intersects circle $1$ at point $C$. Prove that $AB = BC$.
(V. Prasovov, Moscow)
The diagonals $AC$ and $BD$ of the cyclic quadrilateral $ABCD$ intersect at a point O. It is known that $\angle BAD = 60^o$ and $AO = 3OC$. Prove that the sum of some two sides of a quadrilateral is equal to the sum of the other two sides.
Given the isosceles triangle $ABC$ with $| AB | = | AC | = 10$ and $| BC | = 15$. Let points $P$ in $BC$ and $Q$ in $AC$ chosen such that $| AQ | = | QP | = | P C |$. Calculate the ratio of areas of the triangles $(PQA): (ABC)$.
In triangle $ABC$, $M$ is the intersection point of the medians, $O$ is the center of the inscribed circle. Prove that if the line $OM$ is parallel to the side $BC$, then the point $O$ is equidistant from the sides $AB$ and $AC$.
Let $ABCD$ a convex quadrilateral (this means that all interior angles are smaller than $180^o$), such that there exist a point $M$ on line segment $AB$ and a point $N$ on line segment $BC$ having the property that $AN$ cuts the quadrilateral in two parts of equal area, and such that the same property holds for $CM$.
Prove that $MN$ cuts the diagonal $BD$ in two segments of equal length.
Let the isosceles right triangle $ABC$ with $\angle A= 90^o$ . The points $E$ and $F$ are taken on the ray $AC$ so that $\angle ABE = 15^o$ and $CE = CF$. Determine the measure of the angle $CBF$.
Given an isosceles triangle $ABC$ with $AB = AC$, suppose $D$ is the midpoint of the $AC$. The circumcircle of the $DBC$ triangle intersects the altitude from $A$ at point $E$ inside the triangle $ABC$, and the circumcircle of the triangle $AEB$ cuts the side $BD$ at point $F$. If $CF$ cuts $AE$ at point $G$, prove that $AE = EG$.
In a triangle $[ABC]$, $\angle C = 2\angle A$. A point $D$ is marked on the side $[AC]$ such that $\angle ABD = \angle DBC$. Knowing that $AB = 10$ and $CD = 3$, what is the length of the side $[BC]$?
The line joining the midpoints of two opposite sides of a convex quadrilateral makes equal angles with the diagonals. Show that the diagonals are equal.
Triangle $ABC$ is equilateral. The point $D$ lies on the extension of $AB$ beyond $B$, the point $E$ lies on the extension of $CB$ beyond $B$, and $|CD| = |DE|$. Prove that $|AD| = |BE|$.
[img]https://1.bp.blogspot.com/-QnAXFw3ijn0/XzR0YjqBQ3I/AAAAAAAAMU0/0TvhMQtBNjolYHtgXsQo2OPGJzEYSfCwACLcBGAsYHQ/s0/2015%2BMohr%2Bp3.png[/img]
In the triangle $ABC$ it is known that $2AC=AB$ and $\angle A = 2\angle B$. In this triangle draw the angle bisector $AL$, and mark point $M$, the midpoint of the side $AB$. It turned out that $CL = ML$. Prove that $\angle B= 30^o$.
(Hilko Danilo)
In a convex quadrilateral $ABCD$, $\angle ABC = 90^o$ , $\angle BAC = \angle CAD$, $AC = AD, DH$ is the alltitude of the triangle $ACD$. In what ratio does the line $BH$ divide the segment $CD$?
Let $ABC$ be a triangle and $I$ its incenter. The circle passing through $A, C$ and $I$ intersect the line $BC$ for second time at point $X$. The circle passing through $B, C$ and $I$ intersects the line $AC$ for second time at point $Y$. Show that the segments $AY$ and $BX$ have equal length.