Found problems: 133
Let $ABC$ be a triangle and $D$ a point inside the triangle, located on the median of $A$. Prove that if $\angle BDC = 180^o - \angle BAC$, then $AB \cdot CD = AC \cdot BD$.
In an acute triangle $\triangle ABC$ the points $C_m, A_m, B_m$ are the midpoints of $AB, BC, CA$ respectively. Inside the triangle $\triangle ABC$ a point $P$ is chosen so that $\angle PCB = \angle B_mBC$ and $\angle PAB = \angle ABB_m.$ A line passing through $P$ and perpendicular to $AC$ meets the median $BB_m$ at $E.$ Prove that $E$ lies on the circumcircle of the triangle $\triangle A_mB_mC_m.$
[i](K. Ivanov )[/i]
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.
Restore the triangle with a compass and a ruler given the intersection point of altitudes and the feet of the median and angle bisectors drawn to one side. (No research required.)
From the vertex $C$ in triangle $ABC$, draw a straight line that bisects the median from $A$. In what ratio does this line divide the segment $AB$?
[img]https://1.bp.blogspot.com/-SxWIQ12DIvs/XzcJv5xoV0I/AAAAAAAAMY4/Ezfe8bd7W-Mfp2Qi4qE_gppbh9Fzvb4XwCLcBGAsYHQ/s0/1995%2BMohr%2Bp3.png[/img]
In the triangle $ABC$ the median $AM$ is drawn. Is it possible that the radius of the circle inscribed in $\vartriangle ABM$ could be twice as large as the radius of the circle inscribed in $\vartriangle ACM$ ?
( D . Fomin , Leningrad)
Find the greatest value of $p$ and the smallest value of $q$ such that for any triangle in the plane, the inequality
\[p<\frac{a+m}{b+n}<q\]
holds, where $a,b$ are it's two sides and $m,n$ their corresponding medians.
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 an acute triangle, the distance from the midpoint of any side to the opposite vertex is equal to the sum of the distances from it to sides of the triangle. Prove that this triangle is equilateral.
Let $ABC$ be a triangle with medians $m_a$ , $m_b$, $m_c$. Prove that:
(a) There is a triangle with side lengths $m_a$ ,$m_b$, $m_c$.
(b) This triangle is similar to $ABC$ if and only if the squares of the side lengths of triangle $ABC$ form an arithmetical sequence.
Prove that if two medians in a triangle are equal in length, then the triangle is isosceles.
(Note: A median in a triangle is a segment which connects a vertex of the triangle to the midpoint of the opposite side of the triangle.)
Let $B_1,C_1$ be the midpoints of sides $AC,AB$ of a triangle $ABC$ respectively. The tangents to the circumcircle at $B$ and $C$ meet the rays $CC_1,BB_1$ at points $K$ and $L$ respectively. Prove that $\angle BAK = \angle CAL$.
On sides $AB, BC$ and $CA$ of triangle $ABC$ are located points $P, M$ and $K$, respectively, so that $AM, BK$ and $CP$ intersect in one point and the sum of the vectors $\overrightarrow{AM}, \overrightarrow{BK}$ and $\overrightarrow{CP}$ equals $ \overrightarrow{0}$. Prove that $K, M$ and $P$ are midpoints of the sides of triangle $ABC$ on which they are located.
In acute angled triangle $ABC$ we denote by $a,b,c$ the side lengths, by $m_a,m_b,m_c$ the median lengths and by $r_{b}c,r_{ca},r_{ab}$ the radii of the circles tangents to two sides and to circumscribed circle of the triangle, respectively. Prove that
$$\frac{m_a^2}{r_{bc}}+\frac{m_b^2}{r_{ab}}+\frac{m_c^2}{r_{ab}} \ge \frac{27\sqrt3}{8}\sqrt[3]{abc}$$
Given is $\triangle ABC$ such that $\angle A = 57^o, \angle B = 61^o$ and $\angle C = 62^o$. Which segment is longer: the angle bisector through $A$ or the median through $B$?
(N.Beluhov)
In a triangle $ABC$ with an angle $\angle CAB =30^o$ draw median $CD$. If the formed $\vartriangle ACD$ is isosceles, find tan $\angle DCB$.
Let $a, b$ and $c$ denote the side lengths and $m_a, m_b$ and $m_c$ of the median's lengths in an arbitrary triangle. Show that $$\frac34 < \frac{m_a + m_b + m_c}{a + b + c}<1$$ Also show that there is no narrower range that for each triangle that contains the fraction
$$\frac{m_a + m_b + m_c}{a + b + c}$$
Let a$,b,c$ be sidelengths of a triangle and $m_a,m_b,m_c$ the medians at the corresponding sides. Prove that
$$m_a\left(\frac{b}{a}-1\right)\left(\frac{c}{a}-1\right)+
m_b\left(\frac{a}{b}-1\right)\left(\frac{c}{b}-1\right)
+m_c\left(\frac{a}{c}-1\right)\left(\frac{b}{c}-1\right)\geq 0.$$
(FYROM)
Let $BM$ be a median in an acute-angled triangle $ABC$. A point $K$ is chosen on the line through $C$ tangent to the circumcircle of $\vartriangle BMC$ so that $\angle KBC = 90^\circ$. The segments $AK$ and $BM$ meet at $J$. Prove that the circumcenter of $\triangle BJK$ lies on the line $AC$.
Aleksandr Kuznetsov, Russia
In triangle $ABC$, medians $AA'$, $BB'$, $CC'$ are extended until they intersect with the circumcircle at points $A_0$, $B_0$, $C_0$, respectively. It is known that the intersection point M of the medians of triangle $ABC$ divides the segment $AA_0$ in half. Prove that the triangle $A_0B_0C_0$ is isosceles.
The median $AD$ of triangle $ABC$ intersects its inscribed circle (with center $O$) at the points $X$ and $Y$. Find the angle $XOY$ if $AC = AB + AD$.
(A Fedotov)
Points $C_1, A_1$ and $B_1$ are marked on the sides $AB, BC$ and $CA$ of a triangle $ABC$ so that the segments $AA_1, BB_1$, and $CC_1$ are concurrent (see the fig.). It is known that the area of the white part of the triangle $ABC$ is equal to the area of its black part. Prove that at least one of the segments $AA_1, BB_1, CC_1$ is a median of the triangle $ABC$.
[img]https://1.bp.blogspot.com/-nVVhqdRdf0s/X-WVmt_gyqI/AAAAAAAAM40/943sCRGyCPwT-vqIilTCtXOXHByRLIvPwCLcBGAsYHQ/s0/2014%2Bbelarus%2B11.6.png[/img]
In a triangle $ABC, AM$ is a median, $BK$ is a bisectrix, $L=AM\cap BK$. It is known that $BC=a, AB=c, a>c$.
Given that the circumcenter of triangle $ABC$ lies on the line $CL$, find $AC$
I. Voronovich
The median $AM$ is drawn in the triangle $ABC$ ($AB \ne AC$). The point $P$ is the foot of the perpendicular drawn on the segment $AM$ from the point $B$. On the segment $AM$ we chose such a point $Q$ that $AQ = 2PM$. Prove that $\angle CQM = \angle BAM$.
The medians of the sides $AC$ and $BC$ in the triangle $ABC$ are perpendicular to each other. Prove that $\frac12 <\frac{|AC|}{|BC|}<2$.