Found problems: 351
A (convex) trapezoid $ABCD$ is good, if it is inscribed in a circle, sides $AB$ and $CD$ are the bases and $CD$ is shorter than $AB$. For a good trapezoid $ABCD$ the following terms are defined:
$\bullet$ The parallel to $AD$ passing through $B$ intersects the extension of side $CD$ at point $S$.
$\bullet$ The two tangents passing through $S$ on the circumircle of the trapezoid touch the circle at $E$ and $F$, where $E$ lies on the same side of the straight line $CD$ as $A$.
Give the simplest possible equivalent condition (expressed in side lengths and / or angles of the trapezoid) so that with a good trapezoid $ABCD$ the two angles $\angle BSE$ and $\angle FSC$ have the same measure.
(Walther Janous)
A point $M$ is found inside a convex quadrilateral $ABCD$ such that triangles $AMB$ and $CMD$ are isoceles ($AM = MB, CM = MD$) and $\angle AMB= \angle CMD = 120^o$ . Prove that there exists a point N such that triangles$ BNC$ and $DNA$ are equilateral.
(I.Sharygin)
Show that any triangle can be subdivided into isosceles triangles.
Let be given an isosceles triangle $ABC$ with the base $AB$. A point $P$ is chosen on the altitude $CD$ so that the incircles of $ABP$ and $PECF$ are congruent, where $E$ and $F$ are the intersections of $AP$ and $BP$ with the opposite sides of the triangle, respectively. Prove that the incircles of triangles $ADP$ and $BCP$ are also congruent.
In a right-angled and isosceles triangle, the two catheti are both length $1$. Find the length of the shortest line segment dividing the triangle into two figures with the same area, and specify the location of this line segment
Two different points $X$ and $Y$ are chosen in the plane. Find all the points $Z$ in this plane for which the triangle $XYZ$ is isosceles.
In an isosceles triangle $ABC$ with base $AC$, on side $BC$ is selected point $K$ so that $\angle BAK = 24^o$. On the segment $AK$ the point $M$ is chosen so that $\angle ABM = 90^o$, $AM=2BK$. Find the values of all angles of triangle $ABC$.
Points $E$ and $F$ are chosen on the base side $AD$ and the lateral side $AB$ of an isosceles trapezoid $ABCD$, respectively. Quadrilateral $CDEF$ is an isosceles trapezoid as well. Prove that $AE \cdot ED = AF \cdot FB$.
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.
On the legs $AC$ and $BC$ of an isosceles right-angled triangle with a right angle $C$, points $D$ and $E$ are taken, respectively, so that $CD = CE$. Perpendiculars on line $AE$ from points $C$ and $D$ intersect segment $AB$ at points $P$ and $Q$, respectively. Prove that $BP = PQ$.
Let $ABC$ be an isosceles triangle $ABC$ with base $BC$ insribed in a circle. The segment $AD$ is the diameter of the circle, and point $P$ lies on the smaller arc $BD$. Line $DP$ intersects rays $AB$ and $AC$ at points $M$ and $N$, and the lines $BP$ and $CP$ intersects the line $AD$ at points $Q$ and $R$. Prove that the midpoint of the segment $MN$ lies on the circumscribed circle of triangle $PQR$.
In an isosceles triangle $ABC$ ($AB = BC$), the midline parallel to side $BC$ intersects the incircle at a point $F$ that does not lie on the base $AC$. Prove that the tangent to the circle at point $F$ intersects the bisector of angle $C$ on side $AB$.
Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D$ be a point on the base $BC$ such that $BD:DC = 2: 1$. Note on the segment $AD$ a point $P$ such that $\angle BAC= \angle BPD $. Prove that $\angle BPD = 2 \angle CPD$.
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 $\angle AMD = \angle AND$.
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.
In the isosceles triangle $ABC$, with $AB = AC$, the angle bisector of $\angle B$ intersects side $AC$ at $B'$. Suppose that $ B B' + B'A = BC$. Find the angles of the triangle.
Let $G$ be the centroid of a triangle $ABC$. Prove that if $AB+GC = AC+GB$, then the triangle is isosceles
Vero had an isosceles triangle made of paper. Using scissors, he divided it into three smaller triangles and painted them blue, red and green. Having done so, he observed that:
$\bullet$ with the blue triangle and the red triangle an isosceles triangle can be formed,
$\bullet$ with the blue triangle and the green triangle an isosceles triangle can be formed,
$\bullet$ with the red triangle and the green triangle an isosceles triangle can be formed.
Show what Vero's triangle looked like and how he might have made the cuts to make this situation be possible.
Let $ABC$ be a triangle with circumcircle $\Gamma$. $D$ is the midpoint of arc $BC$ (this arc does not contain $A$). $E$ is the common point of $BC$ and the perpendicular bisector of $BD$. $F$ is the common point of $AC$ and the parallel to $AB$ containing $D$. $G$ is the common point of $EF$ and $AB$. $H$ is the common point of $GD$ and $AC$. Show that $GAH$ is isosceles.
Let $ABCD$ be a cyclic quadrangle, let the diagonals $AC$ and $BD$ cross at $O$, and let $I$ and $J$ be the incentres of the triangles $ABC$ and $ABD$, respectively. The line $IJ$ crosses the segments $OA$ and $OB$ at $M$ and $N$, respectively. Prove that the triangle $OMN$ is isosceles.
At what $n$ can a regular $n$-gon be cut by disjoint diagonals into $n- 2$ isosceles (including equilateral) triangles?
Triangle $ABC$ is right-angled and isosceles with a right angle at the vertex $C$. On rays $CB$ on vertex $B$ is selected point F, on rays $BA$ on vertex $A$ is selected point G so that $AG = BF.$ The ray $GD$ is drawn so that it intersects with ray $AC$ at point $D$ with $\angle FGD = 45^o$. Find $\angle FDG$.
(Bogdan Rublev)
In an isosceles triangle $\vartriangle ABC, \angle ABC = 20^o$ and $BC = AB$. Points $P$ and $Q$ are chosen on sides $BC$ and $AB$, respectively, so that $\angle PAC = 50^o$ and $\angle QCA = 60^o$ . Prove that $\angle PQC = 30^o$ .
Let $ABC$ be a triangle where $|AB| = |AC|$. Points $P$ and $Q$ are different from the vertices of the triangle and lie on the sides $AB$ and $AC$, respectively. Prove that the circumcircle of the triangle $APQ$ passes through the circumcenter of $ABC$ if and only if $|AP| = |CQ|$.
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)