Found problems: 1049
$ABCD$ is a parallelogram, line $DF$ is drawn bisecting $BC$ at $E$ and meeting $AB$ (extended) at $F$ from vertex $C$. Line $CH$ is drawn bisecting side $AD$ at $G$ and meeting $AB$ (extended) at $H$. Lines $DF$ and $CH$ intersect at $I$. If the area of parallelogram $ABCD$ is $x$, find the area of triangle $HFI$ in terms of $x$.
Let $K$, $L$, $M$, $N$ be the midpoints of sides $BC$, $CD$, $DA$, $AB$ respectively of a convex quadrilateral $ABCD$. The common points of segments $AK$, $BL$, $CM$, $DN$ divide each of them into three parts. It is known that the ratio of the length of the medial part to the length of the whole segment is the same for all segments. Does this yield that $ABCD$ is a parallelogram?
Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.
Let $l,m$ be two parallel lines in the plane.
Let $P$ be a fixed point between them.
Let $E,F$ be variable points on $l,m$ such that the angle $EPF$ is fixed to a number like $\alpha$ where $0<\alpha<\frac{\pi}2$.
(By angle $EPF$ we mean the directed angle)
Show that there is another point (not $P$) such that it sees the segment $EF$ with a fixed angle too.
Let $ ABCD $ be a parallelogram with $ BC = 17 $. Let $ M $ be the midpoint of $ \overline{BC} $ and let $ N $ be the
point such that $ DANM $ is a parallelogram. What is the length of segment $ \overline{NC} $?
Let $ABCD$ be a rectangle with $BC=3AB$. Show that if $P,Q$ are the points on side $BC$ with $BP = PQ = QC$, then \[\angle DBC+\angle DPC = \angle DQC.\]
Let $ABC$ be an acute triangle with $\angle BAC>\angle BCA$, and let $D$ be a point on side $AC$ such that $|AB|=|BD|$. Furthermore, let $F$ be a point on the circumcircle of triangle $ABC$ such that line $FD$ is perpendicular to side $BC$ and points $F,B$ lie on different sides of line $AC$. Prove that line $FB$ is perpendicular to side $AC$ .
Let $ABCD$ be an arbitrary quadrilateral. Let squares $ABB_1A_2, BCC_1B_2, CDD_1C_2, DAA_1D_2$ be constructed in the exterior of the quadrilateral. Furthermore, let $AA_1PA_2$ and $CC_1QC_2$ be parallelograms. For any arbitrary point $P$ in the interior of $ABCD$, parallelograms $RASC$ and $RPTQ$ are constructed. Prove that these two parallelograms have two vertices in common.
An equilateral triangle with side length 9 is divided into 81 congruent triangles with segments which are parallel to the sides of the triangle. Prove that it cannot be divided into more than 18 parallelograms with sides 1 and 2.
[I]Proposed by O.Vanyushina[/i]
The point $ M$ is chosen inside parallelogram $ ABCD$. Show that $ \angle MAB$ is congruent to $ \angle MCB$, if and only if $ \angle MBA$ and $ \angle MDA$ are congruent.
Let $ABCD$ be a parallelogram. A circle with diameter $AC$ intersects line $BD$ at points $P$ and $Q$. The perpendicular on $AC$ passing through point $C$, intersects lines $AB$ and $AD$ at points $X$ and $Y$, respectively. Prove that the points $P, Q, X$ and $Y$ lie on the same circle.
Let $ABC$ be a triangle with $\angle CAB > 45^o$ and $\angle CBA > 45^o$. Construct an isosceles right angled triangle $RAB$ with $AB$ as its hypotenuse and $R$ inside $ABC$. Also construct isosceles right angled triangles $ACQ$ and $BCP$ having $AC$ and $BC$ respectively as their hypotenuses and lying entirely outside $ABC$. Show that $CQRP$ is a parallelogram.
On the sides $AB$ and $CD$ of the parallelogram $ABCD$ mark points $E$ and $F$, respectively. On the diagonals $AC$ and $BD$ chose the points $M$ and $N$ so that $EM\parallel BD$ and $FN\parallel AC$. Prove that the lines $AF, DE$ and $MN$ intersect at one point.
(B. Rublev)
Describe a construction of quadrilateral $ABCD$ given:
(i) the lengths of all four sides;
(ii) that $AB$ and $CD$ are parallel;
(iii) that $BC$ and $DA$ do not intersect.
In parallelogram $ABCD$, points $E$ and $F$ on segments $AD$ and $CD$ are such that $\angle BCE=\angle BAF$. Points $K$ and $L$ on segments $AD$ and $CD$ are such that $AK=ED$ and $CL=FD$. Prove that $\angle BKD=\angle BLD$.
Given is trapezoid $ABCD$, $M$ and $N$ being the midpoints of the bases of $AD$ and $BC$, respectively.
a) Prove that the trapezoid is isosceles if it is known that the intersection point of perpendicular bisectors of the lateral sides belongs to the segment $MN$.
b) Does the statement of point a) remain true if it is only known that the intersection point of perpendicular bisectors of the lateral sides belongs to the line $MN$?
A circle can be inscribed in the convex quadrilateral $ ABCD$. The incircle touches the sides $ AB, BC, CD, DA$ at $ A_1, B_1, C_1, D_1$ respectively. The points $ E, F, G, H$ are the midpoints of $ A_1B_1, B_1C_1, C_1D_1, D_1A_1$ respectively. Prove that the quadrilateral $ EFGH$ is a rectangle if and only if $ A, B, C, D$ are concyclic.
Given acute triangle $ABC$ with $AB>AC$, let $M$ be the midpoint of $BC$. $P$ is a point in triangle $AMC$ such that $\angle MAB=\angle PAC$. Let $O,O_1,O_2$ be the circumcenters of $\triangle ABC,\triangle ABP,\triangle ACP$ respectively. Prove that line $AO$ passes through the midpoint of $O_1 O_2$.
Let $ABCD$ be a convex quadrilateral. Assume line $AB$ and $CD$ intersect at $E$, and $B$ lies between $A$ and $E$. Assume line $AD$ and $BC$ intersect at $F$, and $D$ lies between $A$ and $F$. Assume the circumcircles of $\triangle BEC$ and $\triangle CFD$ intersect at $C$ and $P$. Prove that $\angle BAP=\angle CAD$ if and only if $BD\parallel EF$.
Let $ A_{1}A_{2}A_{3}A_{4}A_{5}$ be a convex pentagon, such that
\[ [A_{1}A_{2}A_{3}] \equal{} [A_{2}A_{3}A_{4}] \equal{} [A_{3}A_{4}A_{5}] \equal{} [A_{4}A_{5}A_{1}] \equal{} [A_{5}A_{1}A_{2}].\]
Prove that there exists a point $ M$ in the plane of the pentagon such that
\[ [A_{1}MA_{2}] \equal{} [A_{2}MA_{3}] \equal{} [A_{3}MA_{4}] \equal{} [A_{4}MA_{5}] \equal{} [A_{5}MA_{1}].\]
Here $ [XYZ]$ stands for the area of the triangle $ \Delta XYZ$.
Let $E$ be the point of intersection of the diagonals of convex quadrilateral $ABCD$, and let $P,Q,R,$ and $S$ be the centers of the circles circumscribing triangles $ABE,$ $BCE$, $CDE$, and $ADE$, respectively. Then
$\textbf{(A) }PQRS\text{ is a parallelogram}$
$\textbf{(B) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rhombus}$
$\textbf{(C) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a rectangle}$
$\textbf{(D) }PQRS\text{ is a parallelogram if an only if }ABCD\text{ is a parallelogram}$
$\textbf{(E) }\text{none of the above are true}$
In the square $ABCD$ denote by $M$ the midpoint of the side $[AB]$, with $P$ the projection of point $B$ on the line $CM$ and with $N$ the midpoint of the segment $[CP]$, Bisector of the angle $DAN$ intersects the line $DP$ at point $Q$. Show that the quadrilateral $BMQN$ is a parallelogram.
Let $D$ be an interior point of triangle $ABC$. Lines $BD$ and $CD$ intersect sides $AC$ and $AB$ at points $E$ and $F$, respectively. Points $X$ and $Y$ are on the plane such that $BFEX$ and $CEFY$ are parallelograms. Suppose lines $EY$ and $FX$ intersect at a point $T$ inside triangle $ABC$. Prove that points $B$, $C$, $E$, and $F$ are concyclic if and only if $\angle BAD = \angle CAT$.
Prove that if a lattice parallelogram contains at most three lattice points in addition to its vertices, then those are on one of the diagonals.
Consider the parallelogram $ABCD$. A circle is drawn that passes through $A$ and intersects side $AD$ at $N$, side $AB$ at $M$ and diagonal $AC$ in $P$ such that points $A, M, N, P$ are different. Prove that $$AP\cdot AC = AM \cdot AB + AN \cdot AD.$$