Found problems: 1049
I centered incircle of triangle $ABC$ $(m(\hat{B})=90^\circ)$ touches $\left[AB\right], \left[BC\right], \left[AC\right]$ respectively at $F, D, E$. $\left[CI\right]\cap\left[EF\right]={L}$ and $\left[DL\right]\cap\left[AB\right]=N$. Prove that $\left[AI\right]=\left[ND\right]$.
$ABCD$ is a cyclic quadrilateral with $AC \perp BD$; $AC$ meets $BD$ at $E$. Prove that \[ EA^2 + EB^2 + EC^2 + ED^2 = 4 R^2 \]
where $R$ is the radius of the circumscribing circle.
The sides $AB, BC, CD$ and $DA$ of the convex quadrilateral $ABCD$ have midpoints $E, F, G$ and $H$. Prove that the triangles $EFB, FGC, GHD$ and $HEA$ can be put together into a parallelogram equal to $EFGH$.
Quadrilateral $ABCD$ has $AB=BC=CD$, $\angle ABC=70^\circ$, and $\angle BCD=170^\circ$. What is the degree measure of $\angle BAD$?
$ \textbf{(A)}\ 75\qquad
\textbf{(B)}\ 80\qquad
\textbf{(C)}\ 85\qquad
\textbf{(D)}\ 90\qquad
\textbf{(E)}\ 95$
A circle $ \omega$ with center $ O$ is tangent to the rays of an angle $ BAC$ at $ B$ and $ C$. Point $ Q$ is taken inside the angle $ BAC$. Assume that point $ P$ on the segment $ AQ$ is such that $ AQ\perp OP$. The line $ OP$ intersects the circumcircles $ \omega_{1}$ and $ \omega_{2}$ of triangles $ BPQ$ and $ CPQ$ again at points $ M$ and $ N$. Prove that $ OM \equal{} ON$.
[b]Problem 1.[/b] Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, of a triangle $\triangle ABC$ such that $DE\parallel BC$. The circumcircle $k$ of triangle $\triangle ADE$ intersects the lines $BE$ and $CD$ at the points $M$ and $N$ (different from $E$ and $D$). The lines $AM$ and $AN$ intersect the side $BC$ at points $P$ and $Q$ such that $BC=2\cdot PQ$ and the point $P$ lies between $B$ and $Q$. Prove that the circle $k$ passes through the point of intersection of the side $BC$ and the angle bisector of $\angle BAC$.
[i]Nikolai Nikolov[/i]
Let $ABCD$ be a parallelogram and let $K$ and $L$ be points on the segments $BC$ and $CD$, respectively, such that $BK\cdot AD=DL\cdot AB$. Let the lines $DK$ and $BL$ intersect at $P$. Show that $\measuredangle DAP=\measuredangle BAC$.
The diagonals AC and BD of a cyclic ABCD intersect at E. Let O be circumcentre of ABCD. If midpoints of AB, CD, OE are collinear prove that AD=BC.
Bomb
[color=red][Moderator edit: The problem is wrong. See also http://www.mathlinks.ro/Forum/viewtopic.php?t=53090 .][/color]
Given the parallelogram $ABCD$. The circle $S_1$ passes through the vertex $C$ and touches the sides $BA$ and $AD$ at points $P_1$ and $Q_1$, respectively. The circle $S_2$ passes through the vertex $B$ and touches the side $DC$ at points $P_2$ and $Q_2$, respectively. Let $d_1$ and $d_2$ be the distances from $C$ and $B$ to the lines $P_1Q_1$ and $P_2Q_2$, respectively.
Find all possible values of the ratio $d_1:d_2$.
[i](I. Voronovich)[/i]
Let $ABC$ be triangle with the symmedian point $L$ and circumradius $R$. Construct parallelograms $ ADLE$, $BHLK$, $CILJ$ such that $D,H \in AB$, $K, I \in BC$, $J,E \in CA$ Suppose that $DE$, $HK$, $IJ$ pairwise intersect at $X, Y,Z$. Prove that inradius of $XYZ$ is $\frac{R}{2}$ .
Let $ABCD$ be a parallelogram with $\angle ABC=120^\circ$, $AB=16$ and $BC=10$. Extend $\overline{CD}$ through $D$ to $E$ so that $DE=4$. If $\overline{BE}$ intersects $\overline{AD}$ at $F$, then $FD$ is closest to
$\textbf{(A) }1\qquad
\textbf{(B) }2\qquad
\textbf{(C) }3\qquad
\textbf{(D) }4\qquad
\textbf{(E) }5$
[asy]
size(200);
defaultpen(linewidth(0.8));
pair A=origin,B=(16,0),C=(26,10*sqrt(3)),D=(10,10*sqrt(3)),E=(0,10*sqrt(3));
draw(A--B--C--E--B--A--D);
label("$A$",A,S);
label("$B$",B,S);
label("$C$",C,N);
label("$D$",D,N);
label("$E$",E,N);
label("$F$",extension(A,D,B,E),W);
label("$4$",(D+E)/2,N);
label("$16$",(8,0),S);
label("$10$",(B+C)/2,SE);
[/asy]
On the sides of a triangle $ABC$ are constructed similar triangles $ABD,BCE,CAF$ with $k=AD/DB=BE/EC=CF/FA$ and $\alpha=\angle ADB=\angle BEC=\angle CFA$. Prove that the midpoints of the segments $AC,BC,CD$ and $EF$ form a parallelogram with an angle $\alpha$ and two sides whose ratio is $k$.
Let $ABC$ a triangle and $P$ a point such that the triangles $PAB, PBC, PCA$ have the same area and the same perimeter. Prove that if:
a) $P$ is in the interior of the triangle $ABC$ then $ABC$ is equilateral.
b) $P$ is in the exterior of the triangle $ABC$ then $ABC$ is right angled triangle.
A special kind of parallelogram tile is made up by attaching the legs of two right isosceles triangles of side length $1$. We want to put a number of these tiles on the floor of an $n\times n$ room such that the distance from each vertex of each tile to the sides of the room is an integer and also no two tiles overlap. Prove that at least an area $n$ of the room will not be covered by the tiles.
[i]Proposed by Ali Khezeli[/i]
Let $ABCD$ be a cyclic quadrilateral. The lines $BC$ and $AD$ meet at a point $P$. Let $Q$ be the point on the line $BP$, different from $B$, such that $PQ=BP$. Consider the parallelograms $CAQR$ and $DBCS$. Prove that the points $C,Q,R,S$ lie on a circle.
Let $ABC$ be an acute triangle, $H$ its orthocentre, $D$ a point on the side $[BC]$, and $P$ a point such that $ADPH$ is a parallelogram.
Show that $\angle BPC > \angle BAC$.
$ABCD$ is a parallelogram. $P$ is a point outside the parallelogram such that angles $\angle PAB$ and $\angle PCB$ have the same value but opposite orientation. Show that $\angle APB = \angle DPC$.
Let $ABCD$ be a convex quadrilateral with $m(\widehat{ADC}) = 90^{\circ}$. The line through $D$ which is parallel to $BC$ meets $AB$ at $E$. If $m(\widehat{DAC}) = m(\widehat{DAE})$, $|AB|=3$ and $|AC|=4$, then $|AE| = ?$
$\textbf{(A)}\ \frac56 \qquad\textbf{(B)}\ \frac13 \qquad\textbf{(C)}\ \frac12 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ \frac34$
Let $ABCD$ be a parallelogram and $M$ is the intersection of $AC$ and $BD$. The point $N$ is inside of the $\triangle AMB$ such that $\angle AND=\angle BNC$. Prove that $\angle MNC=\angle NDA$ and $\angle MND=\angle NCB$.
Let $AC$ and $BD$ be two chords of a circle with center $O$ such that they intersect at right angles inside the circle at the point $M$. Suppose $K$ and $L$ are midpoints of the chords $AB$ and $CD$ respectively. Prove that $OKML$ is a parallelogram.
Let $O, H$ be circumcenter and orthogonal center of triangle $ABC$. $M,N$ are midpoints of $BH$ and $CH$. $BB'$ is diagonal of circumcircle. If $HONM$ is a cyclic quadrilateral, prove that $B'N=\frac12AC$.
a) The triangle $ABC$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the triangle $A_1B_1C_1$. The corresponding segments $[AB]$ and $[A_1B_1]$ intersect in the point $C_2, [BC]$ and $[B_1C_1]$ -- $A_2, [AC]$ and $[A_1C_1]$ -- $B_2$. Prove that the triangle $A_2B_2C_2$ is similar to the triangle $ABC$.
b) The quadrangle $ABCD$ was turned around the centre of the circumscribed circle by the angle less than $180$ degrees and thus was obtained the quadrangle $A_1B_1C_1D_1$. Prove that the points of intersection of the corresponding lines ( $(AB$) and $(A_1B_1), (BC)$ and $(B_1C_1), (CD)$ and $(C_1D_1), (DA)$ and $(D_1A_1)$ ) are the vertices of the parallelogram.
Let $a,b,c,$ and $d$ be the lengths of sides $MN,NP,PQ,$ and $QM$, respectively, of quadrilateral $MNPQ$. If $A$ is the area of $MNPQ$, then
$\textbf{(A) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is convex}$
$\textbf{(B) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$
$\textbf{(C) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$
$\textbf{(D) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$
$\textbf{(E) }A\ge\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$
Let $ABC$ be an acute triangle. Let $D$ and $E$ be the feet of the altitudes of the triangle $ABC$ from $A$ and $B$, respectively. Let $F$ be the reflection of the point $A$ over $BC$. Let $G$ be a point such that the quadrilateral $ABCG$ is a parallelogram. Show that the circumcircles of triangles $BCF$ , $ACG$ and $CDE$ are concurrent on a point different from $C$.
Given convex quadrilateral $ABCD$ inscribed in circle $(O,R)$ (with center $O$ and radius $R$). With centers the vertices of the quadrilateral and radii $R$, we consider the circles $C_A(A,R), C_B(B,R), C_C(C,R), C_D(D,R)$. Circles $C_A$ and $C_B$ intersect at point $K$, circles $C_B$ and $C_C$ intersect at point $L$, circles $C_C$ and $C_D$ intersect at point $M$ and circles $C_D$ and $C_A$ intersect at point $N$ (points $K,L,M,N$ are the second common points of the circles given they all pass through point $O$). Prove that quadrilateral $KLMN$ is a parallelogram.