Found problems: 351
Let $ABC$ be an isosceles triangle with vertex $A$ and $AB> BC$. Let $k$ be the circle with center $A$ passsing through $B$ and $C$. Let $H$ be the second intersection of $k$ with the altitude of the triangle $ABC$ through $B$. Further let $G$ be the second intersection of $k$ with the median through $B$ in triangle $ABC$. Let $X$ be the intersection of the lines $AC$ and $GH$. Show that $C$ is the midpoint of $AX$.
The point $P$ lies inside $\vartriangle ABC$ so that $\vartriangle BPC$ is isosceles, and angle $P$ is a right angle. Furthermore both $\vartriangle BAN$ and $\vartriangle CAM$ are isosceles with a right angle at $A$, and both are outside $\vartriangle ABC$. Show that $\vartriangle MNP$ is isosceles and right-angled.
[img]https://1.bp.blogspot.com/-i9twOChu774/XzcBLP-RIXI/AAAAAAAAMXA/n5TJCOJypeMVW28-9GDG4st5C47yhvTCgCLcBGAsYHQ/s0/2005%2BMohr%2Bp3.png[/img]
The vertices and the circumcenter of an isosceles triangle lie on four different sides of a square. Find the angles of this triangle.
(I. Bogdanov, B. Frenkin)
Consider a semicircle of center $O$ and diameter $[AB]$, and let $C$ be an arbitrary point on the segment $(OB)$. The perpendicular to the line $AB$ through $C$ intersects the semicircle in $D$. A circle centered in $P$ is tangent to the arc $BD$ in $F$ and to the segments $[AB]$ and $[CD]$ in $G$ and $E$, respectively. Prove that the triangle $ADG$ is isosceles.
Let $I_a$ be the point of the center of an ex-circle of the triangle $ABC$, which touches the side $BC$ . Let $W$ be the intersection point of the bisector of the angle $\angle A$ of the triangle $ABC$ with the circumcircle of the triangle $ABC$. Perpendicular from the point $W$ on the straight line $AB$, intersects the circumcircle of $ABC$ at the point $P$. Prove, that if the points $B, P, I_a$ lie on the same line, then the triangle $ABC$ is isosceles.
(Mykola Moroz)
Prove that any triangle can be cut by at most into $3$ parts, from which an isosceles triangle is formed.
In the acute isosceles triangle $ABC$ the altitudes $BB_1$ and $CC_1$ are drawn, which intersect at the point $H$. Let $L_1$ and $L_2$ be the feet of the angle bisectors of the triangles $B_1AC_1$ and $B_1HC_1$ drawn from vertices $A$ and $H$, respectively. The circumscribed circles of triangles $AHL_1$ and $AHL_2$ intersects the line $B_1C_1$ for the second time at points $P$ and $Q$, respectively. Prove that points $B, C, P$ and $Q$ lie on the same circle.
(M. Plotnikov, D. Hilko)
Two squares and an isosceles triangle are positioned as shown in the figure (the up left vertex of the large square lies on the side of the triangle). Prove that points $A, B$ and $C$ are collinear.
[img]https://cdn.artofproblemsolving.com/attachments/d/c/03515e40f74ced1f8243c11b3e610ef92137ac.png[/img]
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 $AB = AC$.
Consider the isosceles right triangle $ABC$ ($AB = AC$) and the points $M, P \in [AB]$ so that $AM = BP$. Let $D$ be the midpoint of the side $BC$ and $R, Q$ the intersections of the perpendicular from $A$ on$ CM$ with $CM$ and $BC$ respectively. Prove that
a) $\angle AQC = \angle PQB$
b) $\angle DRQ = 45^o$
On the sides $AB$, $BC$ and $CA$ of the isosceles triangle $ABC$ with the vertex at the point $B$ marked the points $M$, $D$ and $K$ respectively so that $AM = 2DC$ and $\angle AMD = \angle KDC$. Prove that $MD = KD$.
In triangle $ABC$, bisectors are drawn $AA_1$ and $CC_1$. Prove that if the length of the perpendiculars drawn from the vertex $B$ on lines $AA1$ and $CC_1$ are equal, then $\vartriangle ABC$ is isosceles.
In the isosceles triangle $ABC$, $AB = AC$. Let $E$ be on side $AB$ such that $\angle ACE = \angle ECB = 18^o$, and let $D$ be the midpoint of side $CB$. If we know the length of $AD$ is $3$ units, what is the length of $CE$?
Points $A,B,C$ with $AB = BC$ are given on a circle with radius $r$, and $D$ is a point inside the circle such that the triangle $BCD$ is equilateral. The line $AD$ meets the circle again at $E$. Show that $DE = r$.
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.)
Consider an isosceles triangle $KL_1L_2$ with $|KL_1|=|KL_2|$ and let $KA, L_1B_1,L_2B_2$ be its angle bisectors. Prove that $\cos \angle B_1AB_2 < \frac35$
Given a regular $2013$-sided polygon, how many isosceles triangles are there whose vertices are vertices vertex of given polygon and haave an angle greater than $120^o$?
Nguyen Tien Lam, High School for Natural Science,Hanoi National University.
Given is an acute-angled triangle $ABC$ in which holds $|BC|: |AC| = 3:$2. Let $D$ be the midpoint of the side $\overline{AC}$, and P the midpoint of the segment $\overline{BD}$. A point $X$ is given on the line $AC$ so that $|AX| = |BC|$, where $A$ is between $X$ and $C$. The line $XP$ intersects the side $\overline{BC}$ at point $E$. The line $DE$ intersects the line $AP$ at point $Y$. Prove that the points $A$, $X$, $Y$, $E$ lie on one circle if and only if $|AB| = |BC|$.
It is given isosceles triangle $ABC$ ($AB=AC$) such that $\angle BAC=108^{\circ}$. Angle bisector of angle $\angle ABC$ intersects side $AC$ in point $D$, and point $E$ is on side $BC$ such that $BE=AE$. If $AE=m$, find $ED$
Prove that among any $51$ vertices of the $101$-regular polygon there are three that are the vertices of an isosceles triangle.
Find all isosceles triangles that cannot be cut into three isosceles triangles with the same sides.
In the quadrilateral $ABCD$ is $AD \parallel BC$ and the angles $\angle A$ and $\angle D$ are acute. The diagonals intersect in $P$. The circumscribed circles of $\vartriangle ABP$ and $\vartriangle CDP$ intersect the line $AD$ again at $S$ and $T$ respectively. Call $M$ the midpoint of $[ST]$. Prove that $\vartriangle BCM$ is isosceles.
[img]https://1.bp.blogspot.com/-C5MqC0RTqwY/Xy1fAavi_aI/AAAAAAAAMSM/2MXMlwb13McCYTrOHm1ZzWc0nkaR1J6zQCLcBGAsYHQ/s0/flanders%2B2016%2Bp1.png[/img]
In a triangle $ABC$ with an angle $\angle CAB =30^o$ draw median $CD$. If the formed $\vartriangle ACD$ is isosceles, find tan $\angle DCB$.
In a certain triangle, the bisectors of the two interior angles were extended to the intersection with the circumscribed circle and two equal chords were obtained. Is it true that the triangle is isosceles?
It is known that $ \angle BAC$ is the smallest angle in the triangle $ ABC$. The points $ B$ and $ C$ divide the circumcircle of the triangle into two arcs. Let $ U$ be an interior point of the arc between $ B$ and $ C$ which does not contain $ A$. The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AU$ at $ V$ and $ W$, respectively. The lines $ BV$ and $ CW$ meet at $ T$.
Show that $ AU \equal{} TB \plus{} TC$.
[i]Alternative formulation:[/i]
Four different points $ A,B,C,D$ are chosen on a circle $ \Gamma$ such that the triangle $ BCD$ is not right-angled. Prove that:
(a) The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AD$ at certain points $ W$ and $ V,$ respectively, and that the lines $ CV$ and $ BW$ meet at a certain point $ T.$
(b) The length of one of the line segments $ AD, BT,$ and $ CT$ is the sum of the lengths of the other two.