### Euler Olympiad 2009-19 IX (Russia) 21p

geometry problems from Euler Olympiads (final stage, from Russia)
with aops links in the names
a Junior Competition

collected inside aops here

2009 - 2019

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 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 $\angle KAD = \angle KCD$.

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 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?

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 $PB+PC<AD$.

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 $BED$.

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 $ABC$.

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 $\angle KBC = 2 \angle KBA$.

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 $AOD$.

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 $NMK$.

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 $2LN = AC$.

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 $BDE$.

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 $AD<2$

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 $AXY$.

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 $AKD$.

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 $SA = SR.$
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 $\angle ADC = 90^\circ.$