Given an acute triangle $ABC$. with $H$ as its orthocenter, lines $\ell_1$ and $\ell_2$ go through $H$ and are perpendicular to each other. Line $\ell_1$ cuts $BC$ and the extension of $AB$ on $D$ and $Z$ respectively. Whereas line $\ell_2$ cuts $BC$ and the extension of $AC$ on $E$ and $X$ respectively. If the line through $D$ and parallel to $AC$ and the line through $E$ parallel to $AB$ intersects at $Y$, prove that $X,Y,Z$ are collinear.

Angle $A$ in triangle $ABC$ is equal to $120^o$. Prove that the distance from the center of the circumscribed circle to the orthocenter is equal to $AB + AC$. (V. Protasov)

Let $\Delta ABC$ be an acute triangle and $H$ its orthocenter. $B_1$ and $C_1$ are the feet of heights from $B$ and $C$, $M$ is the midpoint of $AH$. Point $K$ is on the segment $B_1C_1$, but isn't on line $AH$. Line $AK$ intersects the lines $MB_1$ and $MC_1$ in $E$ and $F$, the lines $BE$ and $CF$ intersect at $N$. Prove that $K$ is the orthocenter of $\Delta NBC$.

Let $ABC$ be a triangle with $H$ its orthocenter. The circle with diameter $[AC]$ cuts the circumcircle of triangle $ABH$ at $K$. Prove that the point of intersection of the lines $CK$ and $BH$ is the midpoint of the segment $[BH]$

Given an acute-angled triangle $ABC$, in which $P$, $M$, $N$ are the midpoints of the sides $AB$, $BC$, $AC$, respectively. A point $H$ is taken inside the triangle and perpendiculars $HK$, $HS$, $HQ$ are lowered from it to the sides $AB$, $BC$, $AC$, respectively ($K \in AB$, $S \in BC$, $Q \in AC$). It turned out that $MK = MQ$, $NS = NK$, $PS=PQ$. Prove that $H$ is the point of intersection of the altitudes of triangle $ABC$.

[b]5.[/b] In triangle $ABC$, let $r_A$ be the line that passes through the midpoint of $BC$ and is perpendicular to the internal bisector of $\angle{BAC}$. Define $r_B$ and $r_C$ similarly. Let $H$ and $I$ be the orthocenter and incenter of $ABC$, respectively. Suppose that the three lines $r_A$, $r_B$, $r_C$ define a triangle. Prove that the circumcenter of this triangle is the midpoint of $HI$.

The altitudes of the acute-angled triangle $ABC$ intersect at the point $H$. On the segments $BH$ and $CH$, the points $B_1$ and $C_1$ are marked, respectively, so that $B_1C_1 \parallel BC$. It turned out that the center of the circle $\omega$ circumscribed around the triangle $B_1HC_1$ lies on the line $BC$. Prove that the circle $\Gamma$, which is circumscribed around the triangle $ABC$, is tangent to the circle $\omega$ .

Find the locus of the points of intersection of the altitudes of the triangles inscribed in a given circle.

[u]Version for Nordic Countries[/u] Six pine trees grow on the shore of a circular lake. It is known that a treasure is submerged at the mid-point $T$ between the intersection points of the altitudes of two triangles, the vertices of one being at three of the $6$ pines, and the vertices of the second one at the other three pines. At how many points $T$ must one dive to find the treasure? [u]Version for Tropical Countries[/u] A captain finds his way to Treasure Island, which is circular in shape. He knows that there is treasure buried at the midpoint of the segment joining the orthocentres of triangles $ABC$ and $DEF$, where $A$, $B$, $C$, $D$, $E$ and $F$ are six palm trees on the shore of the island, not necessarily in cyclic order. He finds the trees all right, but does not know which tree is denoted by which letter. What is the maximum number of points at which the captain has to dig in order to recover the treasure? (S Markelov)

Given an acute triangle $ABC$, $(c)$ its circumcircle with center $O$ and $H$ the orthocenter of the triangle $ABC$. The line $AO$ intersects $(c)$ at the point $D$. Let $D_1, D_2$ and $H_2, H_3$ be the symmetrical points of the points $D$ and $H$ with respect to the lines $AB, AC$ respectively. Let $(c_1)$ be the circumcircle of the triangle $AD_1D_2$. Suppose that the line $AH$ intersects again $(c_1)$ at the point $U$, the line $H_2H_3$ intersects the segment $D_1D_2$ at the point $K_1$ and the line $DH_3$ intersects the segment $UD_2$ at the point $L_1$. Prove that one of the intersection points of the circumcircles of the triangles $D_1K_1H_2$ and $UDL_1$ lies on the line $K_1L_1$.

In the triangle $ABC$, the orthocenter $H$ lies on the inscribed circle. Is this triangle necessarily isosceles?

Let $\vartriangle ABC$ be an acute triangle with altitudes $AA_1, BB_1, CC_1$ and orthocenter $H$. Let $K, L$ be the midpoints of $BC_1, CB_1$. Let $\ell_A$ be the external angle bisector of $\angle BAC$. Let $\ell_B, \ell_C$ be the lines through $B, C$ perpendicular to $\ell_A$. Let $\ell_H$ be the line through $H$ parallel to $\ell_A$. Prove that the centers of the circumcircles of $\vartriangle A_1B_1C_1, \vartriangle AKL$ and the rectangle formed by $\ell_A, \ell_B, \ell_C, \ell_H$ lie on the same line.

Let $I$ be the incenter of a scalene triangle $\vartriangle ABC$. In other words, $\overline{AB},\overline{BC},\overline{CA}$ are distinct. Prove that if $D,E$ are two points on rays $\overrightarrow{BA},\overrightarrow{CA}$, satisfying $\overline{BD}=\overline{CA},\overline{CE}=\overline{BA}$ then line $DE$ pass through the orthocenter of $\vartriangle BIC$. [img]http://2.bp.blogspot.com/-aHCD5tL0FuA/XnYM1LoZjWI/AAAAAAAALeE/C6hO9W9FGhcuUP3MQ9aD7SNq5q7g_cY9QCK4BGAYYCw/s1600/imoc2019g1.png[/img]

Let ${ABC}$ be a non-equilateral triangle. The incircle is tangent to the sides ${BC,CA,AB}$ at ${A_1,B_1,C_1}$, respectively, and M is the orthocenter of triangle ${A_1B_1C_1}$. Prove that ${M}$ lies on the line through the incenter and circumcenter of ${\vartriangle ABC}$.

The point $O$ is the center of the circle circumscribed of the acute triangle $ABC$, and $H$ is the point of intersection of the heights of this triangle. Let $A_1, B_1, C_1$ be the points diametrically opposed to the vertices $A, B , C$ respectively of the triangle, and $A_2, B_2, C_2$ be the midpoints of the segments $[AH], [BH] ¸[CH]$ respectively . Prove that the lines $A_1A_2, B_1B_2, C_1C_2$ are concurrent .

$H$ is the intersection point of the heights $AA'$ and $BB'$ of the acute-angled triangle $ABC$. A straight line, perpendicular to $AB$, intersects these heights at points $D$ and $E$, and side $AB$ at point $P$. Prove that the orthocenter of the triangle $DEH$ lies on segment $CP$.

Altitudes $AH_A, BH_B, CH_C$ of triangle $ABC$ intersect at $H$, and let $M$ be the midpoint of the side $AC$. The bisector $BL$ of $\triangle ABC$ intersects $H_AH_C$ at point $K$. The line through $L$ parallel to $HM$ intersects $BH_B$ in point $T$. Prove that $TK = TL$. [i]Proposed by Anton Trygub[/i]

Let $ABC$ be a triangle and $D, E, F$ be the foots of altitudes drawn from $A,B,C$ respectively. Let $H$ be the orthocenter of $ABC$. Lines $EF$ and $AD$ intersect at $G$. Let $K$ the point on circumcircle of $ABC$ such that $AK$ is a diameter of this circle. $AK$ cuts $BC$ in $M$. Prove that $GM$ and $HK$ are parallel.

In convex pentagon $ABCDE, AC$ is parallel to $DE, AB$ is perpendicular to $AE$, and $BC$ is perpendicular to $CD$. If $H$ is the orthocentre of triangle $ABC$ and $M$ is the midpoint of segment $DE$, prove that $AD, CE$ and $HM$ are concurrent.

Let $ABC$ an acutangle triangle with orthocenter $H$ and circumcenter $O$. Let $D$ the intersection of $AO$ and $BH$. Let $P$ be the point on $AB$ such that $PH=PD$. Prove that the points $B, D, O$ and $P$ lie on a circle.

Points $P$ and $Q$ lies inside triangle $ABC$ such that $\angle ACP =\angle BCQ$ and $\angle CAP = \angle BAQ$. Denote by $D,E$, and $F$ the feet of perpendiculars from $P$ to lines $BC,CA$, and $AB$, respectively. Prove that if $\angle DEF = 90^o$, then $Q$ is the orthocenter of triangle $BDF$.

As shown in figure , points $D,E,F$ lies the sides $AB$, $BC$ , $CA$ of the acute angle $\vartriangle ABC$ respectively. If $\angle EDC = \angle CDF$, $\angle FEA=\angle AED$, $\angle DFB =\angle BFE$, prove that the $CD$, $AE$, $BF$ are the altitudes of $\vartriangle ABC$. [img]https://cdn.artofproblemsolving.com/attachments/3/d/5ddf48e298ad1b75691c13935102b26abe73c1.png[/img]

A straight line passing through the center of the circumscribed circle and the intersection point of the heights of the non-equilateral triangle $ABC$ divides its perimeter and area in the same ratio.Find this ratio.

Triangle $ABC$ is given. An arbitrary circle with center $J$, passing through $B$ and $C$, intersects the sides $AC$ and $AB$ at $E$ and $F$, respectively. Let $X$ be a point such that triangle $FXB$ is similar to triangle $EJC$ (with the same order) and the points $X$ and $C$ lie on the same side of the line $AB$. Similarly, let $Y$ be a point such that triangle $EYC$ is similar to triangle $FJB$ (with the same order) and the points $Y$ and $B$ lie on the same side of the line $AC$. Prove that the line $XY$ passes through the orthocenter of the triangle $ABC$. [i]Proposed by Nguyen Van Linh - Vietnam[/i]

Given an acute $\vartriangle ABC$ whose orthocenter is denoted by $H$. A line $\ell$ passes $H$ and intersects $AB,AC$ at $P ,Q$ such that $H$ is the mid-point of $P,Q$. Assume the other intersection of the circumcircle of $\vartriangle ABC$ with the circumcircle of $\vartriangle APQ$ is $X$. Let $C'$ is the symmetric point of $C$ with respect to $X$ and $Y$ is the another intersection of the circumcircle of $\vartriangle ABC$ and $AO$, where O is the circumcenter of $\vartriangle APQ$. Show that $CY$ is tangent to circumcircle of $\vartriangle BCC'$. [img]https://1.bp.blogspot.com/-itG6m1ipAfE/XndLDUtSf7I/AAAAAAAALfc/iZahX6yNItItRSXkDYNofR5hKApyFH84gCK4BGAYYCw/s1600/2018%2Bimoc%2Bg3.png[/img]