Found problems: 1049
Let $E$ be the midpoint of the side $AB$ of the parallelogram $ABCD$. Let $F$ be the projection of $B$ on $AC$. Prove that the triangle $ABF$ is isosceles
In a triangle $ABC$ with $AB<AC<BC$, the perpendicular bisectors of $AC$ and $BC$ intersect $BC$ and $AC$ at $K$ and $L$, respectively. Let $O$, $O_1$, and $O_2$ be the circumcentres of triangles $ABC$, $CKL$, and $OAB$, respectively. Prove that $OCO_1O_2$ is a parallelogram.
Let $ABCD$ be a convex quadrilateral and $K,L,M,N$ be points on $[AB],[BC],[CD],[DA]$, respectively. Show that,
\[
\sqrt[3]{s_{1}}+\sqrt[3]{s_{2}}+\sqrt[3]{s_{3}}+\sqrt[3]{s_{4}}\leq 2\sqrt[3]{s}
\]
where $s_1=\text{Area}(AKN)$, $s_2=\text{Area}(BKL)$, $s_3=\text{Area}(CLM)$, $s_4=\text{Area}(DMN)$ and $s=\text{Area}(ABCD)$.
Let $ABCD$ be a parallelogram. $E$ and $F$ are on $BC, CD$ respectively such that the triangles $ABE$ and $BCF$ have the same area. Let $BD$ intersect $AE, AF$ at $M, N$ respectively. Prove there exists a triangle whose side lengths are $BM, MN, ND$.
Consider a parallelogram $[ABCD]$ such that $\angle DAB$ is an acute angle. Let $G$ be a point in line $AB$ different from $B$ such that $\overline{BC}=\overline{GC}$, and let $H$ be a point in line $BC$ different from $B$ such that $\overline{AB}=\overline{AH}$. Prove that triangle $[GDH]$ is isosceles.
Two circles $\mathcal{C}_1$ and $\mathcal{C}_2$ intersect in $P$ and $Q$. A line through $P$ intersects $\mathcal{C}_1$ and $\mathcal{C}_2$ again at $A$ and $B$, respectively, and $X$ is the midpoint of $AB$. The line through $Q$ and $X$ intersects $C_1$ and $C_2$ again at $Y$ and $Z$, respectively. Prove that $X$ is the midpoint of $YZ$.
A parallelogram $ABCD$ with $|AD| =|BD|$ has been given. A point $E$ lies on line segment $|BD|$ in such a way that $|AE| = |DE|$. The (extended) line $AE$ intersects line segment $BC$ in $F$. Line $DF$ is the angle bisector of angle $CDE$. Determine the size of angle $ABD$.
[asy]
unitsize (3 cm);
pair A, B, C, D, E, F;
D = (0,0);
A = dir(250);
B = dir(290);
C = B + D - A;
E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, D);
F = extension(A, E, B, C);
draw(A--B--C--D--cycle);
draw(A--F--D--B);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, NE);
dot("$D$", D, NW);
dot("$E$", E, S);
dot("$F$", F, SE);
[/asy]
Let $H$ be the orthocenter of an acute trangle $ABC$ with circumcircle $\Gamma$. Let $P$ be a point on the arc $BC$ (not containing $A$) of $\Gamma$, and let $M$ be a point on the arc $CA$ (not containing $B$) of $\Gamma$ such that $H$ lies on the segment $PM$. Let $K$ be another point on $\Gamma$ such that $KM$ is parallel to the Simson line of $P$ with respect to triangle $ABC$. Let $Q$ be another point on $\Gamma$ such that $PQ \parallel BC$. Segments $BC$ and $KQ$ intersect at a point $J$. Prove that $\triangle KJM$ is an isosceles triangle.
$ABCD$ is a parallelogram. The excircle of $ABC$ opposite $A$ has center $E$ and touches the line $AB$ at $X$. The excircle of $ADC$ opposite $A$ has center $F$ and touches the line $AD$ at $Y$. The line $FC$ meets the line$ AB$ at $W$, and the line $EC$ meets the line $AD$ at $Z$. Show that $WX = YZ$.
A quadrilateral $ABCD$ is inscribed in a circle with center $O$. Let $A_1 B_1 C_1 D_1$ be the image of $ABCD$ after rotation with center $O$ and angle $\alpha \in (0,90^\circ)$. The points $P,Q,R$ and $S$ are intersections of $AB$ and $A_1 B_1$, $BC$ and $B_1 C_1$, $CD$ and $C_1 D_1$, and $DA$ and $D_1 A_1$. Prove that $PQRS$ is a parallelogram.
Let $ABCD$ be a parallelogram. Construct the equilateral triangle $DCE$ on the side $DC$ and outside of parallelogram. Let $P$ be an arbitrary point in plane of $ABCD$. Show that
\[PA+PB+AD \geq PE.\]
$D$ is an arbitrary point lying on side $BC$ of $\triangle{ABC}$. Circle $\omega_1$ is tangent to segments $AD$ , $BD$ and the circumcircle of $\triangle{ABC}$ and circle $\omega_2$ is tangent to segments $AD$ , $CD$ and the circumcircle of $\triangle{ABC}$. Let $X$ and $Y$ be the intersection points of $\omega_1$ and $\omega_2$ with $BC$ respectively and take $M$ as the midpoint of $XY$. Let $T$ be the midpoint of arc $BC$ which does not contain $A$. If $I$ is the incenter of $\triangle{ABC}$, prove that $TM$ goes through the midpoint of $ID$.
Show that the four perpendiculars dropped from the midpoints of the sides of a cyclic quadrilateral to the respective opposite sides are concurrent.
[b]Note by Darij:[/b] A [i]cyclic quadrilateral [/i]is a quadrilateral inscribed in a circle.
$ABC$ is a triangle, and $E$ and $F$ are points on the segments $BC$ and $CA$ respectively, such that $\frac{CE}{CB}+\frac{CF}{CA}=1$ and $\angle CEF=\angle CAB$. Suppose that $M$ is the midpoint of $EF$ and $G$ is the point of intersection between $CM$ and $AB$. Prove that triangle $FEG$ is similar to triangle $ABC$.
Let $P$ be a point inside a triangle $ABC$ such that $\angle PAC= \angle PCB$. Let the projections of $P$ onto $BC$, $CA$, and $AB$ be $X,Y,Z$ respectively. Let $O$ be the circumcenter of $\triangle XYZ$, $H$ be the foot of the altitude from $B$ to $AC$, $N$ be the midpoint of $AC$, and $T$ be the point such that $TYPO$ is a parallelogram. Show that $\triangle THN$ is similar to $\triangle PBC$.
[i]Proposed by Sammy Luo[/i]
Let convex quadrilateral $ ABCD$ be inscribed in a circle centers at $ O.$ The opposite sides $ BA,CD$ meet at $ H$, the diagonals $ AC,BD$ meet at $ G.$ Let $ O_{1},O_{2}$ be the circumcenters of triangles $ AGD,BGC.$ $ O_{1}O_{2}$ intersects $ OG$ at $ N.$ The line $ HG$ cuts the circumcircles of triangles $ AGD,BGC$ at $ P,Q$, respectively. Denote by $ M$ the midpoint of $ PQ.$ Prove that $ NO \equal{} NM.$
Let $P$ be a point inside a triangle $ABC$, and $A_1B_2,B_1C_2,C_1A_2$ be segments passing through $P$ and parallel to $AB, BC, CA$ respectively, where points $A_1, A_2$ lie on $BC, B_1, B_2$ on $CA$, and $C_1,C_2$ on $AB$. Prove that
\[ \text{Area}(A_1A_2B_1B_2C_1C_2) \ge \frac{1}{2}\text{Area}(ABC)\]
Let $ABCD$ be a quadrilateral, let $O$ be the intersection point of diagonals $AC$ and $BD$, and let $P$ be the intersection point of sides $AB$ and $CD$. Consider the parallelograms $AODE$ and $BOCF$. Prove that $E, F$ and $P$ are collinear.
In a parallelogram $ABCD$, let $M$ be the point on the $BC$ side such that $MC = 2BM$ and let $N$ be the point of side $CD$ such that $NC = 2DN$. If the distance from point $B$ to the line $AM$ is $3$, calculate the distance from point $N$ to the line $AM$.
Convex quadrilateral $ ABCD$ has $ AB\equal{}9$ and $ CD\equal{}12$. Diagonals $ AC$ and $ BD$ intersect at $ E$, $ AC\equal{}14$, and $ \triangle AED$ and $ \triangle BEC$ have equal areas. What is $ AE$?
$ \textbf{(A)}\ \frac{9}{2}\qquad \textbf{(B)}\ \frac{50}{11}\qquad \textbf{(C)}\ \frac{21}{4}\qquad \textbf{(D)}\ \frac{17}{3}\qquad \textbf{(E)}\ 6$
Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$.
[i]Proposed by Carlos Yuzo Shine, Brazil[/i]
The diagonals of the cyclic quadrilateral $ ABCD$ are intersecting at the point $ E$.
$ K$ and $ M$ are the midpoints of $ AB$ and $ CD$, respectively. Let the points $ L$ on $ BC$ and $ N$ on $ AD$ s.t.
$ EL\perp BC$ and $ EN\perp AD$.Prove that $ KM\perp LN$.
A triangle $ ABC$ has an obtuse angle at $ B$. The perpindicular at $ B$ to $ AB$ meets $ AC$ at $ D$, and $ |CD| \equal{} |AB|$.
Prove that $ |AD|^2 \equal{} |AB|.|BC|$ if and only if $ \angle CBD \equal{} 30^\circ$.
Let be given a parallelogram $ ABCD$ and two points $ A_1$, $ C_1$ on its sides $ AB$, $ BC$, respectively. Lines $ AC_1$ and $ CA_1$ meet at $ P$. Assume that the circumcircles of triangles $ AA_1P$ and $ CC_1P$ intersect at the second point $ Q$ inside triangle $ ACD$. Prove that $ \angle PDA \equal{} \angle QBA$.
Point $P$ lies inside of parallelogram $ABCD$. Perpendicular lines to $PA,PB,PC$ and $PD$ through $A,B,C$ and $D$ construct convex quadrilateral $XYZT$. Prove that $S_{XYZT}\geq 2S_{ABCD}$.
[i]Proposed by Siamak Ahmadpour[/i]