The point $O$ inside the convex polygon makes isosceles triangle with all the pairs of its vertices. Prove that $O$ is the centre of the circumscribed circle. [u]other formulation:[/u] $P$ is a convex polygon and $X$ is an interior point such that for every pair of vertices $A, B$, the triangle $XAB$ is isosceles. Prove that all the vertices of $P$ lie on a circle with center $X$.

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?

An isosceles triangle $ABC$ with base $AC$ is given. On the rays $CA$, $AB$ and $BC$, the points $D, E$ and $F$ were marked, respectively, in such a way that $AD = AC$, $BE = BA$ and $CF = CB$. Find the sum of the angles $\angle ADB$, $\angle BEC$ and $\angle CFA$.

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 $|AB| = |BC|$.

Anton has an isosceles right triangle, which he wants to cut into $9$ triangular parts in the way shown in the picture. What is the largest number of the resulting $9$ parts that can be equilateral triangles? A more formal description of partitioning. Let triangle $ABC$ be given. We choose two points on its sides so that they go in the order $AC_1C_2BA_1A_2CB_1B_2$, and no two coincide. In addition, the segments $C_1A_2$, $A_1B_2$ and $B_1C_2$ must intersect at one point. Then the partition is given by segments $C_1A_2$, $A_1B_2$, $B_1C_2$, $A_1C_2$, $B_1A_2$ and $C_1B_2$. [img]https://cdn.artofproblemsolving.com/attachments/0/5/5dd914b987983216342e23460954d46755d351.png[/img]

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]

In the triangle $ABC$, the bisector of angle $\angle B$ meets $AC$ at $D$ and the bisector of angle $\angle C$ meets $AB$ at $E$. The bisectors meet each other at $O$. Furthermore, $OD = OE$. Prove that either $ABC$ is isosceles or $\angle BAC = 60^\circ$.

Show that if a triangle has two excircles of the same size, then the triangle is isosceles. (Note: The excircle $ABC$ to the side $ a$ touches the extensions of the sides $AB$ and $AC$ and the side $BC$.)

Let a square $\mathbf P=P_1P_2P_3P_4$ be given in the plane. Determine the locus of all vertices $A$ of isosceles triangles $ABC,AB=BC$ such that the vertices $B,C$ are points of the square $\mathbf P.$

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.

An isosceles triangle $ABC$ with base $AC$ is given. On the rays $CA$, $AB$ and $BC$, the points $D, E$ and $F$ were marked, respectively, in such a way that $AD = AC$, $BE = BA$ and $CF = CB$. Find the sum of the angles $\angle ADB$, $\angle BEC$ and $\angle CFA$.

Let $ABC$ be an acute triangle with midpoint $M$ of $AB$. The point $D$ lies on the segment $MB$ and $I_1, I_2$ denote the incenters of $\triangle ADC$ and $\triangle BDC$. Given that $\angle I_1MI_2=90^{\circ}$, show that $CA=CB$.

Let $ABC$ be an isosceles right-angled triangle. Point $D$ is chosen on the prolongation of the hypothenuse $AB$ beyond point $A$ so that $AB = 2AD$. Points $M$ and $N$ on side $AC$ satisfy the relation $AM = NC$. Point $K$ is chosen on the prolongation of $CB$ beyond point $B$ so that $CN = BK$. Determine the angle between lines $NK$ and $DM$. (M.Kungozhin)

Given a triangle $ABC$. We construct two angle bisectors in the corners $A$ and $B$. Than we construct two lines parallel to those ones through the point $C$. $D$ and $E$ are intersections of those lines with the bisectors. It happens, that $(DE)$ line is parallel to $(AB)$. Prove that the triangle is isosceles.

In an isosceles trapezoid $ABCD$, the base $AB$ is twice as large as the base $CD$. Point $M$ is the midpoint of $AB$. It is known that the center of the circle inscribed in the triangle $MCB$ lies on the circle circumscribed around the triangle $MDC$. Find the angle $\angle MBC$. [img]https://cdn.artofproblemsolving.com/attachments/8/a/7af6a1d32c4e2affa49cb3eed9c10ba1e7ab71.png[/img]

In an isosceles triangle $ABC$ with $AB = AC$ and $\angle A = 30^o$, points $L$ and $M$ lie on the sides $AB$ and $AC$, respectively such that $AL = CM$. Point $K$ lies on $AB$ such that $\angle AMK = 45^o$. If $\angle LMC = 75^o$, prove that $KM +ML = BC$. [i]Proposed by Mahdi Etesamifard - Iran[/i]

Let $ABC$ be an isosceles triangle ($AB=AC$). The altitude $AH$ and the perpendiculare bisector $(e)$ of side $AB$ intersect at point $M$ . The perpendicular on line $(e)$ passing through $M$ intersects $BC$ at point $D$. If the circumscribed circle of the triangle $BMD$ intersects line $(e)$ at point $S$ , the prove that: a) $BS // AM$ . b) quadrilateral $AMBS$ is rhombus.

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 $\cos \angle B_1AB_2 < \frac35$

In a triangle $ABC$, right in $A$ and isosceles, let $D$ be a point on the side $AC$ ($A \ne D \ne C$) and $E$ be the point on the extension of $BA$ such that the triangle $ADE$ is isosceles. Let $P$ be the midpoint of segment $BD$, $R$ be the midpoint of the segment $CE$ and $Q$ the intersection point of $ED$ and $BC$. Prove that the quadrilateral $ARQP$ is a square.

In an isosceles right-angled triangle $ABC$, the right angle is at $A$. $D$ lies so on the side $BC$ that $2CD = DB$. Let $E$ be the projection of $B$ onto $AD$. What is the measure fof angle $\angle CED $?

The triangle $ABC$ is isosceles and has a right angle at the vertex $A$. Construct all points that simultaneously satisfy the following two conditions: (i) are equidistant from points $A$ and $B$ (ii) heve distance exactly three times from point $C$ as far as from point $B$.

An isosceles right-angled triangle $ABC$ is given. On the extensions of sides $AB$ and $AC$, behind vertices $B$ and $C$ equal segments $BK$ and $CL$ were laid. $E$ and F are the points of intersection of the segment $KL$ and the lines perpendicular to the $KC$ , passing through the points $B$ and $A$, respectively. Prove that $EF = FL$.

In an isosceles triangle $ABC, I$ is the center of the inscribed circle, $M_1$ is the midpoint of the side $BC, K_2, K_3$ are the points of contact of the inscribed circle of the triangle with segments $AC$ and $AB$, respectively. The point $P$ lies on the circumcircle of the triangle $BCI$, and the angle $M_1PI$ is right. Prove that the lines $BC, PI, K_2K_3$ intersect at one point. (Mikhail Plotnikov)

Let $P$ be a point inside the triangle $ABC$ such that $AB = BC$, $\angle ABC = 80^o$, $\angle PAC = 40^o$ and $\angle ACP = 30^o$. Find $\angle BPC$. (G Galperin)

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]