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

Let $ABCD$ be a parallelogram. Let $G,H$ be the feet of the altitudes from $A$ to $CD$ and $BC$ respectively. If $AD=15$, $AG=12$, and $AH=16$, find the length of $AB$. [i]2019 CCA Math Bonanza Lightning Round #2.4[/i]

In trapezoid $ ABCD$ with bases $ AB$ and $ CD$, we have $ AB\equal{}52$, $ BC\equal{}12$, $ CD\equal{}39$, and $ DA\equal{}5$. The area of $ ABCD$ is [asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W);[/asy] $ \text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

Compute the maximum area of a rectangle which can be inscribed in a triangle of area $M$.

Two circles are internally tangent at $N$. The chords $BA$ and $BC$ of the larger circle are tangent to the smaller circle at $K$ and $M$ respectively. $Q$ and $P$ are midpoint of arcs $AB$ and $BC$ respectively. Circumcircles of triangles $BQK$ and $BPM$ are intersect at $L$. Show that $BPLQ$ is a parallelogram.

In an acute triangle $\triangle{ABC}$, let $AE$ and $BF$ be highs of it, and $H$ its orthocenter. The symmetric line of $AE$ with respect to the angle bisector of $\sphericalangle{A}$ and the symmetric line of $BF$ with respect to the angle bisector of $\sphericalangle{B}$ intersect each other on the point $O$. The lines $AE$ and $AO$ intersect again the circuncircle to $\triangle{ABC}$ on the points $M$ and $N$ respectively. Let $P$ be the intersection of $BC$ with $HN$; $R$ the intersection of $BC$ with $OM$; and $S$ the intersection of $HR$ with $OP$. Show that $AHSO$ is a paralelogram.

A cube with side length $ 1$ is sliced by a plane that passes through two diagonally opposite vertices $ A$ and $ C$ and the midpoints $ B$ and $ D$ of two opposite edges not containing $ A$ and $ C$, ac shown. What is the area of quadrilateral $ ABCD$? [asy]import three; size(200); defaultpen(fontsize(8)+linewidth(0.7)); currentprojection=obliqueX; dotfactor=4; draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4")); draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1)); draw((0.5,1,0)--(0,1,0)--(0,1,1)); dot((0.5,0,0)); label("$A$",(0.5,0,0),WSW); dot((0,1,1)); label("$C$",(0,1,1),NE); dot((0.5,1,0.5)); label("$D$",(0.5,1,0.5),ESE); dot((0,0,0.5)); label("$B$",(0,0,0.5),NW);[/asy]$ \textbf{(A)}\ \frac {\sqrt6}{2} \qquad \textbf{(B)}\ \frac {5}{4} \qquad \textbf{(C)}\ \sqrt2 \qquad \textbf{(D)}\ \frac {3}{2} \qquad \textbf{(E)}\ \sqrt3$

In a parallelogram $ABCD$, the bisector of $\angle A$ intersects $BC$ at $M$ and the extension of $DC$ at $N$. Let $O$ be the circumcircle of the triangle $MCN$. Prove that $\angle OBC = \angle ODC$

Let $\Gamma$ be a circunference and $O$ its center. $AE$ is a diameter of $\Gamma$ and $B$ the midpoint of one of the arcs $AE$ of $\Gamma$. The point $D \ne E$ in on the segment $OE$. The point $C$ is such that the quadrilateral $ABCD$ is a parallelogram, with $AB$ parallel to $CD$ and $BC$ parallel to $AD$. The lines $EB$ and $CD$ meets at point $F$. The line $OF$ cuts the minor arc $EB$ of $\Gamma$ at $I$. Prove that the line $EI$ is the angle bissector of $\angle BEC$.

Consider two triangles, $ABC$ and $DEF$, and any point $O$. We take any point $X$ in $\vartriangle ABC$ and any point $Y$ in $\vartriangle DEF$ and draw a parallelogram $OXY Z$. Prove that the locus of all possible points $Z$ form a polygon. How many sides can it have? Prove that its perimeter is equal to the sum of perimeters of the original triangles.

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.

In a parallelogram $ABCD$ , $BD$ is the largest diagonal. By matching $B$ with $D$ by a bend, a regular pentagon is formed. Calculate the measures of the angles formed by the diagonal $BD$ with each of the sides of the parallelogram.

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]

It is given parallelogram $ABCD$. On it's sides $AB, BC, CD, DA$ are chosen points $E, F, G, H$ such that area of $EFGH$ is half of the area of $ABCD$. Show that at least one of the quadrilaterals $ABFH$ and $AEGD$ is parallelogram.

The sides of the parallelogram serve as the diagonals of the four squares. The vertices of the squares lying in the part of the plane external to the parallelogram (the sides of the squares emerging from these vertices do not have common points with the parallelogram) serve as the vertices of a quadrilateral of area $a$, the four vertices opposite to them form a quadrilateral of area $b$. Find the area of the parallelogram.

Let $B$ and $C$ be two fixed points on a circle centered at $O$ that are not diametrically opposed. Let $A$ be a variable point on the circle distinct from $B$ and $C$ and not belonging to the perpendicular bisector of $BC$. Let $H$ be the orthocenter of $\triangle ABC$, and $M$ and $N$ be the midpoints of the segments $BC$ and $AH$, respectively. The line $AM$ intersects the circle again at $D$, and finally, $NM$ and $OD$ intersect at $P$. Determine the locus of points $P$ as $A$ moves around the circle.

In the parallelogram $ABCD$, a line through $C$ intersects the diagonal $BD$ at $E$ and $AB$ at $F$. If $F$ is the midpoint of $AB$ and the area of $\vartriangle BEC$ is $100$, find the area of the quadrilateral $AFED$.

Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$. (1) Prove that $F,B,C,E$ are concyclic. (2) Denote $A_{1}$ the centre of the circle passing through $F,B,C,E$. $B_{1}$, $C_{1}$ are difined similarly. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.

The bisector of angle $BAD$ in the parallelogram $ABCD$ intersects the lines $BC$ and $CD$ at the points $K$ and $L$ respectively. It is known that $ABCD$ is not a rhombus. Prove that the centre of the circle passing through the points $C, K$ and $L$ lies on the circle passing through the points $B, C$ and $D$.

Let $ABCD$ be a parallelogram whose diagonals intersect in $M$. Suppose that the circumcircle of $ABM$ intersects the segment $AD$ in a point $E \ne A$ and that the circumcircle of $EMD$ intersects the segment $BE$ in a point $F \ne E$. Show that $\angle ACB=\angle DCF$.

Given triangle $ABC$. Point $M$ is the midpoint of side $BC$, and point $P$ is the projection of $B$ to the perpendicular bisector of segment $AC$. Line $PM$ meets $AB$ in point $Q$. Prove that triangle $QPB$ is isosceles.

Let $O,A,B,C$ be variable points in the plane such that $OA=4$, $OB=2\sqrt3$ and $OC=\sqrt {22}$. Find the maximum value of the area $ABC$. [i]Mihai Baluna[/i]

Let a quadrilateral $ABCD$ have an inscribed circle and let $K, L, M, N$ be the tangency points of the sides $AB, BC, CD$ and $DA$, respectively. Consider the orthocenters of each of the triangles $\vartriangle AKN, \vartriangle BLK, \vartriangle CML$ and $\vartriangle DNM$. Prove that these four points are the vertices of a parallelogram.

Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$. [i]Calvin Deng.[/i]

On the sides $BC$, $CA$, $AB$ of an acute-angled triangle $ABC$, we erect (outwardly) the squares $BB_aC_aC$, $CC_bA_bA$, $AA_cB_cB$, respectively. On the sides $B_cB_a$ and $C_aC_b$ of the triangles $BB_cB_a$ and $CC_aC_b$, we erect (outwardly) the squares $B_cB_vB_uB_a$ and $C_aC_uC_vC_b$. Prove that $B_uC_u\parallel BC$. [i]Comment.[/i] This problem originates in the 68th Moscow MO 2005, and a solution was posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=30184 . However ingenious this solution is, there is a different one which shows a bit more: $B_uC_u=4\cdot BC$. Darij