Found problems: 509
A parallelogram $ABCD$ is given. Two circles with centers at the vertices $A$ and $C$ pass through $B$. The straight line $\ell$ that passes through $B$ and crosses the circles at second time at points $X, Y$ respectively. Prove that $DX = DY$.
Inside the equilateral triangle $ABC$, points $P$ and $Q$ are chosen so that the quadrilateral $APQC$ is convex, $AP = PQ = QC$ and $\angle PBQ = 30^o$. Prove that $AQ = BP$.
Let $ABC$ be an acute triangle such that $\angle BAC = 45^o$. Let $D$ a point on $AB$ such that $CD \perp AB$. Let $P$ be an internal point of the segment $CD$. Prove that $AP\perp BC$ if and only if $|AP| = |BC|$.
In isosceles triangle $ABC$ with $AB=AC$, consider point $D$ on the base $BC$ and point $E$ on side $AC$ such that $ \angle BAD = 2 \angle CDE$. Prove that $AD=AE$.
The equal segments $AB$ and $CD$ intersect at the point $O$ and divide it by the relation $AO: OB = CO: OD = 1: 2 $. The lines $AD$ and $BC$ intersect at the point $M$. Prove that $DM = MB$.
$A, B, C, D, E$ are five points on the same circle, so that $ABCDE$ is convex and we have $AB = BC$ and $CD = DE$. Suppose that the lines $(AD)$ and $(BE)$ intersect at $P$, and that the line $(BD)$ meets line $(CA)$ at $Q$ and line $(CE)$ at $T$. Prove that the triangle $PQT$ is isosceles.
(I. Voronovich)
Let $ABC$ be an isosceles triangle such that $AB = AC$ and $\angle A = 20^o$. Let $M$ be the foot of the altitude from $C$ and let $N$ be a point on the side $AC$ such that $CN =\frac12 BC$. Determine the measure of the angle $AMN$.
On the numerical line, arrange a system of closed segments of length $1$ without common points (endpoints included) so that any infinite arithmetic progression with any non zero difference and any first term has a common point with a segment of the system.
In the convex quadrilateral $ABCD$, point $K$ is the midpoint of $AB$, point $L$ is the midpoint of $BC$, point $M$ is the midpoint of CD, and point $N$ is the midpoint of $DA$. Let $S$ be a point lying inside the quadrilateral $ABCD$ such that $KS = LS$ and $NS = MS$ .Prove that $\angle KSN = \angle MSL$.