Found problems: 200
In a triangle, the ratio of the interior angles is $1 : 5 : 6$, and the longest
side has length $12$. What is the length of the altitude (height) of the triangle that
is perpendicular to the longest side?
The side $AC$ of an acute triangle $ABC$ is the diameter of the circle $c_1$ and side $BC$ is the diameter of the circle $c_2$. Let $E$ be the foot of the altitude drawn from the vertex $B$ of the triangle and $F$ the foot of the altitude drawn from the vertex $A$. In addition, let $L$ and $N$ be the points of intersection of the line $BE$ with the circle $c_1$ (the point $L$ lies on the segment $BE$) and the points of intersection of $K$ and $M$ of line $AF$ with circle $c_2$ (point $K$ is in section $AF$). Prove that $K LM N$ is a cyclic quadrilateral.
Let $ABC$ be an acute triangle with its altitudes $BE,CF$. $M$ is the midpoint of $BC$. $N$ is the intersection of $AM$ and $EF. X$ is the projection of $N$ on $BC$. $Y,Z$ are respectively the projections of $X$ onto $AB,AC$. Prove that $N$ is the orthocenter of triangle $AYZ$.
Nguyễn Minh Hà
Prove that if the orthocenter divides all heights of a triangle in the same proportion, the triangle is equilateral.
A triangle $ABC$ is given with $AC = BC = 5$. The altitude of this triangle drawn from vertex $A$ has length $4$. Calculate the length of the altitude of $ABC$ drawn from vertex $C$.
On triangle $ABC$ the $AM$ ($M\in BC$) is median and $BB_1$ and $CC_1$ ($B_1 \in AC,C_1 \in AB$) are altitudes. The stright line $d$ is perpendicular to $AM$ at the point $A$ and intersect the lines $BB_1$ and $CC_1$ at the points $E$ and $F$ respectively. Let denoted with $\omega$ the circle passing through the points $E, M$ and $F$ and with $\omega_1$ and with $\omega_2$ the circles that are tangent to segment $EF$ and with $\omega$ at the arc $EF$ which is not contain the point $M$. If the points $P$ and $Q$ are intersections points for $\omega_1$ and $\omega_2$ then prove that the points $P, Q$ and $M$ are collinear.
In the acute-angled triangle $ABC$, the altitudes $BP$ and $CQ$ were drawn, and the point $T$ is the intersection point of the altitudes of $\Delta PAQ$. It turned out that $\angle CTB = 90 {} ^ \circ$. Find the measure of $\angle BAC$.
(Mikhail Plotnikov)
Denote in the triangle $ABC$ by $h$ the length of the height drawn from vertex $A$, and by $\alpha = \angle BAC$. Prove that the inequality $AB + AC \ge BC \cdot \cos \alpha + 2h \cdot \sin \alpha$ . Are there triangles for which this inequality turns into equality?
Let $B_1$ be the foot of the altitude from the vertex $B$ in the acute-angled $\triangle ABC$. Let $D$ be the midpoint of side $AB$, and $O$ be the circumcentre of $\triangle ABC$. Line $B_1D$ meets line $CO$ at $E$. Prove that the points $B, C, B_1$, and $E$ lie on a circle.
Show that if $ h_A, h_B,$ and $ h_C$ are the altitudes of $ \triangle ABC$, and $ r$ is the radius of the incircle, then $$ h_A + h_B + h_C \ge 9r$$
Given a cyclic quadrilateral $ABCD$ such that the tangents at points $B$ and $D$ to its circumcircle $k$ intersect at the line $AC$. The points $E$ and $F$ lie on the circle $k$ so that the lines $AC, DE$ and $BF$ parallel. Let $M$ be the intersection of the lines $BE$ and $DF$. If $P, Q$ and $R$ are the feet of the altitides of the triangle $ABC$, prove that the points $P, Q, R$ and $M$ lie on the same circle
In the triangle $ABC$, the bisector of angle $A$ is equal to the half-sum of the height and median drawn from vertex $A$. Prove that if $\angle A$ is obtuse, then $AB = AC$.
In a triangle two altitudes are not smaller than the sides on to which they are dropped. Find the angles of the triangle.
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 $\angle PQ'Q = \angle PQB$.
In the triangle $ABC$, the length of the altitude from $A$ to $BC$ is equal to $1$. $D$ is the midpoint of $AC$. What are the possible lengths of $BD$?
Angle $A$ of the acute-angled triangle $ABC$ equals $60^o$ . Prove that the bisector of one of the angles formed by the altitudes drawn from $B$ and $C$, passes through the circumcircle 's centre.
(V . Pogrebnyak , year 12 student , Vinnitsa,)
The altitueds $A {{A} _ {1}} $, $B {{B} _ {1}}$ and $C {C} _ 1$ are drawn in the acute triangle $ABC$. . The perpendicular $AK$ is drawn from the vertex $A$ on the line ${{A} _ {1}} {{B} _ {1}}$, and the perpendicular $BL$ is drawn from the vertex $B$ on the line ${{C} _ {1}} {{B} _ {1}}$. Prove that ${{A} _ {1}} K = {{B} _ {1}} L$.
(Maria Rozhkova)
In triangle $ABC$, the altitude $AH$ passes through midpoint of the median $BM$. Prove that in the triangle $BMC$ also one of the altitudes passes through the midpoint of one of the medians.
Let $ABC$ be an acute-angled triangle, $M$ the midpoint of the side $AC$ and $F$ the foot on $AB$ of the altitude through the vertex $C$. Prove that $AM = AF$ holds if and only if $\angle BAC = 60^o$.
(Karl Czakler)
Let $a > b > c$ be sides of a triangle and $h_a,h_b,h_c$ be the corresponding altitudes.
Prove that $a+h_a > b+h_b > c+h_c$.
Exists a triangle whose three altitudes have lengths $1, 2$ and $3$ respectively?
In the triangle $ABC$, $AD$ is altitude, $M$ is the midpoint of $BC$. It is known that $\angle BAD = \angle DAM = \angle MAC$. Find the values of the angles of the triangle $ABC$
The triangle $ABC$ has $\angle ACB = 30^o$, $BC = 4$ cm and $AB = 3$ cm . Compute the altitudes of the triangle.
Let $BB_1$ and $CC_1$ be the altitudes of the acute-angled triangle $ABC$. From the point $B_1$ the perpendiculars $B_1E$ and $B_1F$ are drawn on the sides $AB$ and $BC$ of the triangle, respectively, and from the point $C_1$ the perpendiculars $C_1 K$ and $C_1L$ on the sides $AC$ and $BC$, respectively. It turned out that the lines $EF$ and $KL$ are perpendicular. Find the measure of the angle $A$ of the triangle $ABC$.
(Alexander Dunyak)
Suppose the points $D, E, F$ lie on sides $BC, CA, AB$, respectively, so that $AD, BE, CF$ are the altitudes. Also, let $AD$ and $EF$ intersect at $P$. Prove that $$\frac{AP}{AD} \ge 1 - \frac{BC^2}{AB^2 + CA^2}$$