This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

AND
OR
NO

Found problems: 351

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 $
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 $BP = PQ$.
Consider an isosceles triangle $ABC$, where $AB = AC$. $D$ is a point on the $AC$ side and $P$ a point on the segment $BD$ so that the angle $\angle APC = 90^o$ and $ \angle ABP = \angle BCP $. Determine the ratio $AD: DC$.
Let $ABC$ be an isosceles triangle ($AB = AC$), $D$ be the midpoint of $BC$, and $M$ be the midpoint of $AD$. On the segment $BM$ take a point $N$ such that $\angle BND = 90^o$. Find the angle $ANC$.
Let $ABC$ be an isosceles triangle with $AC = BC$ and circumcircle $k$. The point $D$ lies on the shorter arc of $k$ over the chord $BC$ and is different from $B$ and $C$. Let $E$ denote the intersection of $CD$ and $AB$. Prove that the line through $B$ and $C$ is a tangent of the circumcircle of the triangle $BDE$. (Karl Czakler)
Outside the square $ABCD$, the rhombus $BCMN$ is constructed with angle $BCM$ obtuse . Let $P$ be the intersection point of the lines $BM$ and $AN$ . Prove that $DM \perp CP$ and the triangle $DPM$ is right isosceles .
An isosceles triangle $ABC$ with $AB = AC$ and $\angle A = 30^o$ is inscribed in a circle with center $O$. Point $D$ lies on the shorter arc $AC$ so that $\angle DOC = 30^o$, and point $G$ lies on the shorter arc $AB$ so that $DG = AC$ and $AG < BG$. The line $BG$ intersects $AC$ and $AB$ at $E$ and $F$, respectively. (a) Prove that triangle $AFG$ is equilateral. (b) Find the ratio between the areas of triangles $AFE$ and $ABC$.
In a trapezoid, the length of one of the diagonals is equal to the sum of the lengths of the bases, and the angle between the diagonals is $60$ degrees. Prove that this trapezoid is isosceles.
The triangle $ABC$ is equilateral. Find the locus of the points $M$ such that the triangles $ABM$ and $ACM$ are both isosceles.
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.
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 $\angle CFA$.
An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is given. Circles with centers $O_1$ and $O_2$ are inscribed in triangles $ABC$ and $ABD$. Prove that line $O_1O_2$ is perpendicular on $BC$.
Let $\vartriangle ABC$ 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 internal angle bisector of $\angle BEC$. Prove that $|BD|= |EF|$ if and only if $|AF| = |EC|$.
Let $ABC$ be an isosceles triangle with $\angle B = 120^o$ . Points $P$ and $Q$ are chosen on the prolongations of segments $AB$ and $CB$ beyond point $B$ so that the rays $AQ$ and $CP$ intersect and are perpendicular to each other. Prove that $\angle PQB = 2\angle PCQ$. (A.Akopyan, D.Shvetsov)
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?
Given an isosceles triangle $ABC$ with $AB = AC$, suppose $D$ is the midpoint of the $AC$. The circumcircle of the $DBC$ triangle intersects the altitude from $A$ at point $E$ inside the triangle $ABC$, and the circumcircle of the triangle $AEB$ cuts the side $BD$ at point $F$. If $CF$ cuts $AE$ at point $G$, prove that $AE = EG$.
Let $ABCD$ be an isosceles trapezoid, $AD=BC$, $AB \parallel CD$. The diagonals of the trapezoid intersect at the point $O$, and the point $M$ is the midpoint of the side $AD$. The circle circumscribed around the triangle $BCM$ intersects the side $AD$ at the point $K$. Prove that $OK \parallel AB$.
Let $ABC$ be an isosceles triangle $(AB=AC)$ so that $\angle A< 2 \angle B$ . Let $D,Z $ points on the extension of height $AM$ so that $\angle CBD = \angle A$ and $\angle ZBA = 90^\circ$. Let $E$ the orthogonal projection of $M$ on height $BF$, and let $K$ the orthogonal projection of $Z$ on $AE$. Prove that $ \angle KDZ = \angle KDB = \angle KZB$.
A triangle $ABC$ is given with $AC = BC = 5$. The altitude of this triangle drawn from vertex $A$ has length $4$. Calculate the length of the altitude of $ABC$ drawn from vertex $C$.
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 $[BP]$.
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]
Let $ABC$ be an isosceles triangle with base $AB$. We choose a point $P$ inside the triangle on altitude through $C$. The circle with diameter $CP$ intersects the straight line through $B$ and $P$ again at the point $D_P$ and the Straight through $A$ and $C$ one more time at point $E_P$. Prove that there is a point $F$ such that for any choice of $P$ the points $D_P , E_P$ and $F$ lie on a straight line. [i](Walther Janous)[/i]
Let $ABC$ be an isosceles triangle with apex $A$ and altitude $AD$. On $AB$, choose a point $F$ distinct from $B$ such that $CF$ is tangent to the incircle of $ABD$. Suppose that $\vartriangle BCF$ is isosceles. Show that those conditions uniquely determine: a) which vertex of $BCF$ is its apex, b) the size of $\angle BAC$
On the sides of triangle $ABC$, isosceles right-angled triangles $AUB, CVB$, and $AWC$ are placed. These three triangles have their right angles at vertices $U, V$ , and $W$, respectively. Triangle $AUB$ lies completely inside triangle $ABC$ and triangles $CVB$ and $AWC$ lie completely outside $ABC$. See the figure. Prove that quadrilateral $UVCW$ is a parallelogram. [asy] import markers; unitsize(1.5 cm); pair A, B, C, U, V, W; A = (0,0); B = (2,0); C = (1.7,2.5); U = (B + rotate(90,A)*(B))/2; V = (B + rotate(90,C)*(B))/2; W = (C + rotate(90,A)*(C))/2; draw(A--B--C--cycle); draw(A--W, StickIntervalMarker(1,1,size=2mm)); draw(C--W, StickIntervalMarker(1,1,size=2mm)); draw(B--V, StickIntervalMarker(1,2,size=2mm)); draw(C--V, StickIntervalMarker(1,2,size=2mm)); draw(A--U, StickIntervalMarker(1,3,size=2mm)); draw(B--U, StickIntervalMarker(1,3,size=2mm)); draw(rightanglemark(A,U,B,5)); draw(rightanglemark(B,V,C,5)); draw(rightanglemark(A,W,C,5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, N); dot("$U$", U, NE); dot("$V$", V, NE); dot("$W$", W, NW); [/asy]
Triangle $ABC$ is inscribed in a circle. Chords $AM$ and $AN$ intersect side $BC$ at points $K$ and $L$ respectively. Prove that if a circle passes through all of the points $K, L, M$ and $N$, then $ABC$ is an isosceles triangle. (V Zhgun)