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.

Triangle $ABC$ is inscribed in circle with center $O$. Let $P$ be a point on arc $AB$ which does not contain point $C$. Perpendicular from point $P$ on line $BO$ intersects side $AB$ in point $S$, and side $BC$ in $T$. Perpendicular from point $P$ on line $AO$ intersects side $AB$ in point $Q$, and side $AC$ in $R$. (i) Prove that triangle $PQS$ is isosceles (ii) Prove that $\frac{PQ}{QR}=\frac{ST}{PQ}$

Pinocchio claims that he can divide an isoceles triangle into three triangles, any two of which can be put together to form a new isosceles triangle. Is Pinocchio lying? (A Shapovalov)

Show that any triangle can be subdivided into isosceles triangles.

Given an isosceles trapezoid $ABCD$ with base $AB$. The diagonals $AC$ and $BD$ intersect at point $S$. Let $M$ the midpoint of $BC$ and the bisector of the angle $BSC$ intersect $BC$ at $N$. Prove that $\angle AMD = \angle AND$.

In an isosceles triangle $ABC$ ($AB = AC$), the bisector of the angle $B$ intersects $AC$ at point $D$ such that $BC = BD + AD$. Find $\angle A$.

In acute triangle $ABC$, $\angle A = 20^o$. Prove that the triangle is isosceles if and only if $$\sqrt[3]{a^3 + b^3 + c^3 -3abc} = \min\{b, c\}$$, where $a,b, c$ are the side lengths of triangle $ABC$.

The $M$ point is the midpoint of the base $[AC]$ of an isosceles triangle $ABC$. $[MH]$ is orthogonal to $[BC]$ side. Point $P$ is the midpoint of the segment $[MH]$. Prove that $[AH]$ is orthogonal to $[BP]$.

In the isosceles triangle $\vartriangle ABC$ is $AB = AC$. Let $D$ be the midpoint of the segment $BC$. The points $P$ and $Q$ are respectively on the lines $AD$ and $AB$ (with $Q \ne B$) so that $PQ = PC$. Show that $\angle PQC =\frac12 \angle A $

Let $ABC$ be an equilateral $ABC$. Points $M, N, P$ are located on the sides $AC, AB, BC$, respectively, such that $\angle CBM= \frac{1}{2} \angle AMN = \frac{1}{3} \angle BNP$ and $\angle CMP = 90 ^o$. a) Show that $\vartriangle NMB$ is isosceles. b) Determine $\angle CBM$.

Consider the isosceles triangle $ABC$ inscribed in the semicircle of radius $ r$. If the $\vartriangle BCD$ and $\vartriangle CAE$ are equilateral, determine the altitude of $\vartriangle DEC$ on the side $DE$ in terms of $ r$. [img]https://cdn.artofproblemsolving.com/attachments/6/3/772ff9a1fd91e9fa7a7e45ef788eec7a1ba48e.png[/img]

In an isosceles triangle $ABC$ in which $AC = BC$ and $\angle ABC < 60^o$, $I$ and $O$ are the centers of the inscribed and circumscribed circles, respectively. For the triangle $BIO$, the circumscribed circle intersects the side $BC$ again at $D$. Prove that: i) lines $AC$ and $DI$ are parallel, ii) lines $OD$ and $IB$ are perpendicular.

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 triangle $[ABC]$, the bisector in $C$ and the altitude passing through $B$ intersect at point $D$. Point $E$ is the symmetric of point $D$ wrt $BC$ and lies on the circle circumscribed to the triangle $[ABC]$. Prove that the triangle is $[ABC]$ 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 $\angle CFA$.

Let $ABC$ be a triangle with $AB = AC$ ¸ $\angle BAC = 100^o$ and $AD, BE$ angle bisectors. Prove that $2AD <BE + EA$

Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $BF = BE$ and such that $ED$ is the angle bisector of $\angle BEC$. Prove that $BD = EF$ if and only if $AF = EC$.

Given are the points $A$ and $B$ in the plane. If $x$ is a straight line is in that plane, and $x$ does not coincide with the perpendicular bisectror of $AB$, then denote the number of points $C$ located at $x$ such that $\vartriangle ABC$ is isosceles, as the "weight of the line $x$”. Prove that the weight of any line $x$ is at most $5$ and determine the set of points $P$ which has a line with weight $1$, but none with weight $0$.

In triangle $ABC$, the incircle touches the side $AC$ at point $B_1$ and one excircle is touching the same side at point $B_2$. It is known that the segments $BB_1$ and $BB_2$ are equal. Is it true that $\vartriangle ABC$ is isosceles?

Prove that among any $51$ vertices of the $101$-regular polygon there are three that are the vertices of an isosceles triangle.

Angle $C$ of an isosceles triangle $ABC$ equals $120^o$. Each of two rays emitting from vertex $C$ (inwards the triangle) meets $AB$ at some point ($P_i$) reflects according to the rule the angle of incidence equals the angle of reflection" and meets lateral side of triangle $ABC$ at point $Q_i$ ($i = 1,2$). Given that angle between the rays equals $60^o$, prove that area of triangle $P_1CP_2$ equals the sum of areas of triangles $AQ_1P_1$ and $BQ_2P_2$ ($AP_1 < AP_2$).

In triangle $ABC$ the bisector of the angle at $B$ meets $AC$ at $D$ and the bisector of the angle at $C$ meets $AB$ at $E$. These bisectors intersect at $O$ and the lengths of $OD$ and $OE$ are equal. Prove that either $\angle BAC = 60^o$ or triangle $ABC$ is isosceles.

In an isosceles triangle $ABC$ with an angle $\angle A= 20^o$ and base $BC=12$ point $E$ on the side $AC$ is chosen such that $\angle ABE= 30^o$ , and point $F$ on the side $AB$ such that $EF = FC$ . Find the length of $FC$.

Diagonals of an isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ are perpendicular. Let $DE$ be the perpendicular from $D$ to $AB$, and let $CF$ be the perpendicular from $C$ to $DE$. Prove that angle $DBF$ is equal to half of angle $FCD$.

Given an isosceles $ABC$, which has $2AC = AB + BC$. Denote $I$ the center of the inscribed circle, $K$ the midpoint of the arc $ABC$ of the circumscribed circle. Let $T$ be such a point on the line $AC$ that $\angle TIB = 90 {} ^ \circ$. Prove that the line $TB$ touches the circumscribed circle $\Delta KBI$. (Anton Trygub)