Found problems: 200
a) An altitude of a triangle is also a tangent to its circumcircle. Prove that some angle of the triangle is larger than $90^o$ but smaller than $135^o$.
b) Some two altitudes of the triangle are both tangents to its circumcircle. Find the angles of the triangle.
Let $ABC$ be an non- isosceles triangle, $H_a$, $H_b$, and $H_c$ be the feet of the altitudes drawn from the vertices $A, B$, and $C$, respectively, and $M_a$, $M_b$, and $M_c$ be the midpoints of the sides $BC$, $CA$, and $AB$, respectively. The circumscribed circles of triangles $AH_bH_c$ and $AM_bM_c$ intersect for second time at point $A'$. The circumscribed circles of triangles $BH_cH_a$ and $BM_cM_a$ intersect for second time at point $B'$. The circumscribed circles of triangles $CH_aH_b$ and $CM_aM_b$ intersect for second time at point $C'$. Prove that points $A', B'$ and $C'$ lie on the same line.
Let $O$ be the center of the circle circumscribed around $\vartriangle ABC$, let $A_1, B_1, C_1$ be symmetric to $O$ through respective sides of $\vartriangle ABC$. Prove that all altitudes of $\vartriangle A_1B_1C_1$ pass through $O$, and all altitudes of $\vartriangle ABC$ pass through the center of the circle circumscribed around $\vartriangle A_1B_1C_1$.
Let $\triangle{ABC}$ has circumcircle $\Gamma$, drop the perpendicular line from $A$ to $BC$ and meet $\Gamma$ at point $D$, similarly, altitude from $B$ to $AC$ meets $\Gamma$ at $E$. Prove that if $AB=DE, \angle{ACB}=60^{\circ}$
(sorry it is from my memory I can't remember the exact problem, but it means the same)
In a tetrahedron, the midpoints of all the edges lie on the same sphere. Prove that it's altitudes intersect at one point.
Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.
Given an acute triangle $ABC, H$ is the foot of the altitude drawn from the point $A$ on the line $BC, P$ and $K \ne H$ are arbitrary points on the segments $AH$ and$ BC$ respectively. Segments $AC$ and $BP$ intersect at point $B_1$, lines $AB$ and $CP$ at point $C_1$. Let $X$ and $Y$ be the projections of point $H$ on the lines $KB_1$ and $KC_1$, respectively. Prove that points $A, P, X$ and $Y$ lie on one circle.
For the given triangle $ABC$, prove that a point $X$ on the side $AB$ satisfies the condition $\overrightarrow{XA} \cdot\overrightarrow{XB} +\overrightarrow{XC} \cdot \overrightarrow{XC} = \overrightarrow{CA} \cdot \overrightarrow{CB} $, iff $X$ is the basepoint of the altitude or median of the triangle $ABC$.
Let $AA _1$ and $CC_1$ be altitudes of an acute triangle $ABC$. Let $I$ and $J$ be the incenters of the triangles $AA_1C$ and $AC_1C$ respectively. The $C_1J$ and $A_1 I$ lines cut into $T$. Prove that lines $AT$ and $TC$ are perpendicular.
Let $AA_1$, $BB_1$, $CC_1$ be the altitudes of an acute angled triangle $ABC$. On the side $BC$ point $K$ is taken such that $\angle BB_1K=\angle A$. On the side $AB$ a point $M$ is taken such that $\angle BB_1M\angle C$. Let $L$ be the intersection of $BB_1$ and $A_1C_1$. Prove that the quadrilateral $B_1KLM$ is circumscribed.
[I]Proposed by A. Khrabrov, D. Rostovski[/i]
In an acute-angled triangle $ABC$ the altitudes $AU,BV,CW$ intersect at $H$.
Points $X,Y,Z$, different from $H$, are taken on segments $AU,BV$, and $CW$, respectively.
(a) Prove that if $X,Y,Z$ and $H$ lie on a circle, then the sum of the areas of triangles $ABZ, AYC, XBC$ equals the area of $ABC$.
(b) Prove the converse of (a).
Let $ABC$ be an acute triangle with $AB < AC$, and let $BE$ and $CF$ be the altitudes. Let the median $AM$ intersect $BE$ at point $P$, and let line $CP$ intersect $AB$ at point $D$ (see Figure 2). Prove that $DE \parallel BC$, and $AC$ is tangent to the circumcircle of $\vartriangle DEF$.
[img]https://cdn.artofproblemsolving.com/attachments/f/7/bbad9f6019a77c6aa46c3a821857f06233cb93.png[/img]
You are given an acute triangle $ABC$ with circumcircle $\omega$. Points $F$ on $AC$, $E$ on $AB$ and $P, Q$ on $\omega$ are chosen so that $\angle AFB = \angle AEC = \angle APE = \angle AQF = 90^\circ$. Show that lines $BC, EF, PQ$ are concurrent or parallel.
[i]Proposed by Fedir Yudin[/i]
In the trapezoid $ABCD, AB = BC = CD, CH$ is the altitude. Prove that the perpendicular from $H$ on $AC$ passes through the midpoint of $BD$.
In the triangle $ABC$, $\angle B = 2 \angle C$, $AD$ is altitude, $M$ is the midpoint of the side $BC$. Prove that $AB = 2DM$.
Let $ABC$ an acute triangle and $D,E$ and $F$ be the feet of altitudes from $A,B$ and $C$, respectively. The line $EF$ and the circumcircle of $ABC$ intersect at $P$, such that $F$ it´s between $E$ and $P$. Lines $BP$ and $DF$ intersect at $Q$. Prove that if $ED=EP$, then $CQ$ and $DP$ are parallel.
It is given that for no side of the triangle from the height drawn to it, the bisector and the median it is impossible to make a triangle. Prove that one of the angles of the triangle is greater than $135^o$
A median , a bisector and an altitude of a certain triangle intersect at an inner point $O$ . The segment of the bisector from the vertex to $O$ is equal to the segment of the altitude from the vertex to $O$ . Prove that the triangle is equilateral .
Let $H, K, L$ be the feet from the altitudes from vertices $A, B, C$ of the triangle $ABC$, respectively. Prove that
$| AK | \cdot | BL | \cdot| CH | = | HK | \cdot | KL | \cdot | LH | = | AL | \cdot | BH | \cdot | CK | $.
Does there exist a nonisosceles triangle such that the altitude from one vertex, the bisectrix from the second one and the median from the third one are equal?
In an acute-angled triangle $ABC$ on the sides $AB$, $BC$, $AC$ the points $H$, $L$, $K$ so that $CH \perp AB$, $HL \parallel AC$, $HK \parallel BC$. Let $P$ and $Q$ feet of altitudes of a triangle $HBL$, drawn from the vertices $H$ and $B$ respectively. Prove that the feet of the altitudes of the triangle $AKH$, drawn from the vertices $A$ and $H$ lie on the line $PQ$.
If every altitude of a tetrahedron is at least $1$, show that the shortest distance between each pair of opposite edges is more than $2$.
Let $ABC$ be an acute triangle and $D$ be the foot of the altitude from $A$ onto $BC$. A semicircle with diameter $BC$ intersects segments $AB, AC$ and $AD$ in the points $F, E$ and $X$, respectively. The circumcircles of the triangles $DEX$ and $DXF$ intersect $BC$ in $L$ and $N$, respectively, other than $D$. Prove that $BN = LC$.
In a triangle $ABC$, points $H,M,L$ are the feet of the altitude from $C$, the median from $A$, and the angle bisector from $B$, respectively. Show that if triangle $HML$ is equilateral, then so is triangle $ABC$.
In the acute-angled triangle $ABC$, the altitudes $BD$ and $CE$ are drawn. Let $F$ and $G$ be the points of the line $ED$ such that $BF$ and $CG$ are perpendicular to $ED$. Prove that $EF = DG$.