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Dan je šiljastokutan trokut $ABC$ s ortocentrom $H$. Neka je $D$ točka takva da je četverokut $AHCD$ paralelogram. Neka je $p$ okomica na pravac $AB$ kroz polovište $A_1$ stranice $\overline{BC}$. Označimo sjecište pravaca $p$ i $AB$ s $E$, a polovište dužine $\overline{A_1E}$ s $F$. Točku u kojoj paralela s pravcem $BD$ kroz točku $A$ siječe $p$ označimo s $G$. Dokaži da je četverokut $AFA_1C$ tetivan ako i samo ako pravac $BF$ prolazi polovištem dužine $\overline{CG}$.

Neka je $ABC$ šiljastokutan trokut u kojem je $|AB| < |AC|$ te neka je kružnica $k$ sa središtem $O$ njegova opisana kružnica. Neka su $P$ i $Q$ točke redom na stranicama $\overline{BC}$ i $\overline{AB}$ takve da je $AQPO$ paralelogram. Neka su $K$ i $L$ sjecišta simetrale dužine $\overline{OP}$ s kružnicom $k$, pri čemu je $K$ na kraćem luku $\widehat{AB}$. Neka je $M$ drugo sjecište pravca $KQ$ i kružnice $k$. Dokaži da točka $A$ pripada simetrali kuta $\measuredangle QLM$.

Neka je $ABCD$ paralelogram takav da je $|AC| = |BC|$. Neka je $P$ točka na pravcu $AB$ takva da $B$ leži između $A$ i $P$. Opisana kružnica trokuta $ACD$ siječe dužinu $\overline{PD}$ u točki $Q$, $Q \neq D$. Opisana kružnica trokuta $APQ$ siječe dužinu $\overline{PC}$ u točki $R$, $R \neq P$.

Dokaži da se pravci $CD$, $AQ$ i $BR$ sijeku u jednoj točki.

Zadan je konveksan šesterokut $ABCDEF$ kojemu su sveke dvije nasuprotne stranice međusobno različitih duljina i paralelne ($AB \parallel DE$, $BC \parallel EF$ i $CD \parallel FA$). Ako je $|AE| = |BD|$ i $|BF| = |CE|$, dokaži da se šesterokutu $ABCDEF$ može opisati kružnica.

Let $ABCD$ be a parallelogram with side lengths $AB = a$, $AD = 1$, and with $\angle BAD = \alpha$. If $\triangle ABD$ is acute, prove that the four circles of radius 1 with centers $A, B, C, D$ cover the parallelogram if and only if $$a \leq \cos \alpha + \sqrt{3} \sin \alpha.$$

Consider five points $A, B, C, D$ and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral. Let $\ell$ be a line passing through $A$. Suppose that $\ell$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF = EG = EC$. Prove that $\ell$ is the bisector of angle $DAB$.

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 such that $DA = DC$ and $AC$ is the bisector of $\angle DAB$. Point $E$ is chosen such 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 (where $AM \parallel EX$ and $AE \parallel MX$). Prove that lines $BD$, $FX$, and $ME$ are concurrent.

Let $ABCD$ be a parallelogram with $\measuredangle BAD = 60°$ and denote by $E$ the intersection of its diagonals. The circumcircle of the triangle $ACD$ meets the line $BA$ at $K \neq A$, the line $BD$ at $P \neq D$ and the line $BC$ at $L \neq C$. The line $EP$ intersects the circumcircle of the triangle $CEL$ at points $E$ and $M$. Prove that the triangles $KLM$ and $CAP$ are congruent.

The incircle of the triangle $ABC$ touches the sides $BC$, $CA$, and $AB$ in the points $D$, $E$, and $F$, respectively. Let $K$ be the point symmetric to $D$ with respect to the incenter. The lines $DE$ and $FK$ intersect at $S$. Prove that $AS$ is parallel to $BC$.

Let $A$, $B$, $C$, $D$, $E$ be points such that $ABCD$ is a cyclic quadrilateral and $ABDE$ is a parallelogram. The diagonals $AC$ and $BD$ intersect at $S$ and the rays $AB$ and $DC$ intersect at $F$. Prove that $\angle AFS = \angle ECD$.

Let $ABC$ be an acute-angled triangle with $AB < AC$, and let $D$ be the foot of its altitude from $A$. Points $B'$ and $C'$ lie on the rays $AB$ and $AC$, respectively, so that points $B'$, $C'$ and $D$ are collinear and points $B$, $C$, $B'$ and $C'$ lie on one circle with center $O$. Prove that if $M$ is the midpoint of $BC$ and $H$ is the orthocenter of $ABC$, then $DHMO$ is a parallelogram.

Let $ABCD$ be a parallelogram with $\angle DAB < 90^{\circ}$. Let $E \neq B$ be the point on the line $BC$ such that $AE = AB$ and let $F \neq D$ be the point on the line $CD$ such that $AF = AD$. The circumcircle of the triangle $CEF$ intersects the line $AE$ again in $P$ and the line $AF$ again in $Q$. Let $X$ be the reflection of $P$ over the line $DE$ and $Y$ the reflection of $Q$ over the line $BF$. Prove that $A, X$ and $Y$ lie on the same line.

Let $ABC$ be an acute triangle with $AB < AC$. Denote by $D$ the foot of the perpendicular from $A$ to $BC$. Let $E$ be the point such that $ABEC$ is a parallelogram. Let $M$ be a point inside triangle $ABC$ such that $MB = MC$. Let $F$ be the reflection of point $D$ across the tangent to the circumcircle of triangle $ADM$ at point $M$. Prove that $AF = DE$.

Neka su $K$ i $L$ redom polovišta stranica $\overline{CD}$ i $\overline{AD}$ paralelograma $ABCD$. Za točku $T$ unutar paralelograma vrijedi $|KT| = |AK|$ i $|LT| = |CL|$. Neka je $M$ polovište dužine $\overline{BT}$. Dokaži da je $\measuredangle MAT = \measuredangle TCM$.

Neka je $M$ polovište stranice $\overline{BC}$ paralelograma $ABCD$. Ako je $E$ nožište okomice iz točke $D$ na pravac $AM$, dokaži da vrijedi $|CD| = |CE|$.

Neka je $A_1A_2A_3A_4$ konveksan četverokut, $S$ sjecište njegovih dijagonala. Označimo sa $s_k$ površinu trokuta $A_kSA_{k+1}$, ($A_5 = A_1$), $k = 1, 2, 3, 4$. Dokažite da je $$s_2^2 = s_1 s_3 \quad \text{i} \quad 2 s_4 = s_1 + s_3$$ ako i samo ako je $A_1A_2A_3A_4$ paralelogram.

Unutar paralelograma $ABCD$ odabrana je točka $T$ tako da vrijedi $|TC| = |BC|$. Neka su $P$ i $M$ redom polovišta dužina $\overline{CD}$ i $\overline{AT}$. Dokaži da je pravac $BT$ okomit na pravac $PM$.

Nad stranicama trokuta $ABC$ konstruirani su slični trokuti $ABD$, $BCE$, $CAF$ ($k = |AD| : |DB| = |BE| : |EC| = |CF| : |FA|$; $\alpha = \measuredangle ADB = \measuredangle BEC = \measuredangle CFA$). Dokažite da su polovišta dužina $\overline{AC}$, $\overline{BC}$, $\overline{CD}$ i $\overline{EF}$ vrhovi paralelograma, čiji je jedan kut jednak $\alpha$, a omjer duljina odgovarajućih stranica $k$.

U ravnini su dane dvije različite točke $O$ i $P$. Odaberimo paralelogram $ABCD$ kojem je točka $O$ središte. Označimo s $M$ i $N$ redom polovišta dužina $\overline{AP}$ i $\overline{BP}$. Točka $Q$ je presjek dužina $\overline{MC}$ i $\overline{ND}$. Dokažite da točke $O$, $Q$ i $P$ leže na istom pravcu i da točka $Q$ ne ovisi o izboru paralelograma $ABCD$.

Neka je $ABCD$ paralelogram takav da vrijedi $|AB| = 4$, $|AD| = 3$, te je mjera kuta pri vrhu $A$ jednaka $60°$. Kružnica $k_1$ dira stranice $\overline{AB}$ i $\overline{AD}$ dok kružnica $k_2$ dira stranice $\overline{CB}$ i $\overline{CD}$.

Kružnice $k_1$ i $k_2$ su sukladne i dodiruju se izvana. Odredi duljinu polumjera tih kružnica.

Nad stranicama $\overline{AB}$, $\overline{BC}$ trokuta $ABC$ konstruirani su kvadrati $ABKL$, $BCMN$ (koji s trokutom imaju samo zajedničku stranicu).

a) Ako je $D$ točka takva da je $ABCD$ paralelogram, dokaži da su trokuti $ABD$ i $BKN$ sukladni.

b) Dokaži da su polovišta dužina $\overline{AC}$, $\overline{KN}$ i središta kvadrata $ABKL$, $BCMN$ vrhovi kvadrata.