Found problems: 200
Let $AD$ be the altitude $ABC$ of an acute triangle. On the line $AD$ are chosen different points $E$ and $F$ so that $|DE |= |DF|$ and point $E$ is in the interior of triangle $ABC$. The circumcircle of triangle $BEF$ intersects $BC$ and $BA$ for second time at points $K$ and $M$ respectively. The circumcircle of the triangle $CEF$ intersects the $CB$ and $CA$ for the second time at points $L$ and $N$ respectively. Prove that the lines $AD, KM$ and $LN$ intersect at one point.
In triangle $ABC$, points $A_1,B_1,C_1$ are bases of altitudes from vertices $A,B,C$, and points $C_A,C_B$ are the projections of $C_1$ to $AC$ and $BC$ respectively. Prove that line $C_AC_B$ bisects the segments $C_1A_1$ and $C_1B_1$.
Through a point $P$ inside the triangle $ABC$ a line is drawn parallel to the base $AB$, dividing the triangle into two equal areas. If the altitude to $AB$ has a length of $1$, then the distance from $P$ to $AB$ is:
$ \textbf{(A)}\ \frac12 \qquad\textbf{(B)}\ \frac14\qquad\textbf{(C)}\ 2-\sqrt2\qquad\textbf{(D)}\ \frac{2-\sqrt2}{2}\qquad\textbf{(E)}\ \frac{2+\sqrt2}{8} $
Let $AA_1$ and $BB_1$ be the altitudes of acute angled triangle $ABC$. Points $K$ and $M$ are midpoints of $AB$ and $A_1B_1$ respectively. Segments $AA_1$ and $KM$ intersect at point $L$. Prove that points $A$, $K$, $L$ and $B_1$ are noncyclic.
[I]Proposed by S. Berlov[/i]
We consider an acute triangle $ABC$. The altitude from $A$ is $AD$, the altitude from $D$ in triangle $ABD$ is $DE,$ and the altitude from $D$ in triangle $ACD$ is $DF$.
a) Prove that the triangles $ABC$ and $AF E$ are similar.
b) Prove that the segment $EF$ and the corresponding segments constructed from the vertices $B$ and $C$ all have the same length.
In the acute-angled non-isosceles triangle $ABC$, the height $AH$ is drawn. Points $B_1$ and $C_1$ are marked on the sides $AC$ and $AB$, respectively, so that $HA$ is the angle bisector of $B_1HC_1$ and quadrangle $BC_1B_1C$ is cyclic. Prove that $B_1$ and $C_1$ are feet of the altitudes of triangle $ABC$.
In an acute-angled triangle $ABC$, the altitudes $AA_1,BB_1,CC_1$ intersect at point $V$. If the triangles $AC_1V, BA_1V, CB_1V$ have the same area, does it follow that the triangle $ABC$ is equilateral?
In triangle $ABC$ length of altitude $CH$, with $H \in AB$, is equal to half of side $AB$. If $\angle BAC = 45^{\circ}$ find $\angle ABC$
Construct triangle $ABC$, given the altitude and the angle bisector both from $A$, if it is known for the sides of the triangle $ABC$ that $2BC = AB + AC$.
(Alexey Karlyuchenko)
Suppose $ABC$ is an acute triangle. Points $A_1, B_1$ and $C_1$ are chosen on sides $BC, AC$ and $AB$ respectively so that the rays $A_1A, B_1B$ and $C_1C$ are bisectors of triangle $A_1B_1C_1$. Prove that $AA_1, BB_1$ and $CC_1$ are altitudes of triangle $ABC$. (6)
Let $ABC$ be a triangle. Let its altitudes $AD$, $BE$ and $CF$ concur at $H$. Let $K, L$ and $M$ be the midpoints of $BC$, $CA$ and $AB$, respectively. Prove that, if $\angle BAC = 60^o$, then the midpoints of the segments $AH$, $DK$, $EL$, $FM$ are concyclic.
Let $BH_b, CH_c$ be altitudes of an acute-angled triangle $ABC$. The line $H_bH_c$ meets the circumcircle of $ABC$ at points $X$ and $Y$. Points $P,Q$ are the reflections of $X,Y$ about $AB,AC$ respectively. Prove that $PQ \parallel BC$.
[i]Proposed by Pavel Kozhevnikov[/i]
Lengths of sides of triangle $ABC$ are positive integers, and smallest side is equal to $2$. Determine the area of triangle $P$ if $v_c = v_a + v_b$, where $v_a$, $v_b$ and $v_c$ are lengths of altitudes in triangle $ABC$ from vertices $A$, $B$ and $C$, respectively.
In isosceles triangle $ABC$($CA=CB$),$CH$ is altitude and $M$ is midpoint of $BH$.Let $K$ be the foot of the perpendicular from $H$ to $AC$ and $L=BK \cap CM$ .Let the perpendicular drawn from $B$ to $BC$ intersects with $HL$ at $N$.Prove that $\angle ACB=2 \angle BCN$.(M. Kunhozhyn)
An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$. The circle with diameter $AC$ intersects the segment $BE$ at point $P$. A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$. Prove that $\angle PQ'Q = \angle PQB$.
Let $AH$ be an altitude of an equilateral triangle $ABC$. Let $I$ be the incentre of triangle $ABH$, and let $L, K$ and $J$ be the incentres of triangles $ABI,BCI$ and $CAI$ respectively. Determine $\angle KJL$.
Let $AA_1 , BB_1$, and $CC_1$ be the altitudes of the non-isosceles acute-angled triangle $ABC$. The circles circumscibred around the triangles $ABC$ and $A_1 B_1 C$ intersect again at the point $P , Z$ is the intersection point of the tangents to the circumscribed circle of the triangle $ABC$ conducted at points $A$ and $B$ . Prove that lines $AP , BC$ and $ZC_1$ are concurrent.
Denote an acute-angle $\vartriangle ABC $ with sides $a, b, c $ respectively by ${{H}_{a}}, {{H}_{b}}, {{H}_{c}} $ the feet of altitudes ${{h}_{a}}, {{h}_{b}}, {{h}_{c}} $. Prove the inequality:
$$\frac {h_ {a} ^{2}} {{{a} ^{2}} - CH_ {a} ^{2}} + \frac{h_{b} ^{2}} {{{ b}^{2}} - AH_{b} ^{2}} + \frac{h_{c}^{2}}{{{c}^{2}} - BH_{c}^{2}} \ge 3 $$
(Dmitry Petrovsky)
Let $ABC$ be an acute triangle with incircle $\omega$, incenter $I$, and $A$-excircle $\omega_{a}$. Let $\omega$ and $\omega_{a}$ meet $BC$ at $X$ and $Y$, respectively. Let $Z$ be the intersection point of $AY$ and $\omega$ which is closer to $A$. The point $H$ is the foot of the altitude from $A$. Show that $HZ$, $IY$ and $AX$ are concurrent.
[i]Proposed by Nikola Velov[/i]
Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.
In triangle $ABC, AH$ is an altitude ($H$ is on $BC$) and $BE$ is a bisector ($E$ is on $AC$) . We are given that angle $BEA$ equals $45^o$ .Prove that angle $EHC$ equals $45^o$ .
(I. Sharygin , Moscow)
In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$, respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$. Suppose $XP=10$, $PQ=25$, and $QY=15$. The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.
In acute triangle $ABC$, points $D$ and $E$ are the feet of the perpendiculars from $A$ to $BC$ and $B$ to $CA$, respectively. Segment $AD$ is a diameter of circle $\omega$. Circle $\omega$ intersects sides $AC$ and $AB$ at $F$ and $G$ (other than $A$), respectively. Segment $BE$ intersects segments $GD$ and $GF$ at $X$ and $Y$ respectively. Ray $DY$ intersects side $AB$ at $Z$. Prove that lines $XZ$ and $BC$ are perpendicular
In triangle $ABC$ $AD$ and $BE$ are altitudes. Let $L$ be a point on $ED$ such that $ED$ is orthogonal to $BL$. If $LB^2=LD\cdot LE$ prove that triangle $ABC$ is isosceles
Let $ABC$ be an acute-angled triangle with $AB<AC$. Let $D$ be the midpoint of side $BC$ and $BE,CZ$ be the altitudes of the triangle $ABC$. Line $ZE$ intersects line $BC$ at point $O$.
(i) Find all the angles of the triangle $ZDE$ in terms of angle $\angle A$ of the triangle $ABC$
(ii) Find the angle $\angle BOZ$ in terms of angles $\angle B$ and $\angle C$ of the triangle $ABC$