Found problems: 509
Equal line segments are marked in triangle $ABC$. Find its angles.
[img]https://cdn.artofproblemsolving.com/attachments/0/2/bcb756bba15ba57013f1b6c4cbe9cc74171543.png[/img]
Let $ABC$ be an acute-angled triangle with height intersection $H$. Let $G$ be the intersection of parallel of $AB$ through $H$ with the parallel of $AH$ through $B$. Let $I$ be the point on the line $GH$, so that $AC$ bisects segment $HI$. Let $J$ be the second intersection of $AC$ and the circumcircle of the triangle $CGI$. Show that $IJ = AH$
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 a triangle such that $AB = AC$. Let $D$ be a point in $[BC]$ and $E$ a point in $[AD]$ such that
$\angle BE D = \angle BAC = 2 \angle DEC$. Shows that $DB = 2CD$.
[img]https://cdn.artofproblemsolving.com/attachments/d/5/677e19d8e68a89134e17a4ab6051e41f283486.png[/img]
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 $ABC$ be a triangle with $AB=AC$ and $\angle BAC = 40^o$. Points $S$ and $T$ lie on the sides $AB$ and $BC$, such that $\angle BAT = \angle BCS = 10^o$. Lines $AT$ and $CS$ meet at $P$. Prove that $BT = 2PT$.
Let the convex quadrilateral $ABCD$ with $AD = BC$ ¸and $\angle A + \angle B = 120^o$. Take a point $P$ in the plane so that the line $CD$ separates the points $A$ and $P$, and the $DCP$ triangle is equilateral. Show that the triangle $ABP$ is equilateral. It is the true statement for a non-convex quadrilateral?
Let $BC$ and $BD$ be the tangent lines to the circle with diameter $AC$. Let $E$ be the second point of intersection of line $CD$ and the circumscribed circle of triangle $ABC$. Prove that $CD= 2DE$.
(Matthew Kurskyi)
Let $[ABC]$ be a acute-angled triangle and its circumscribed circle $\Gamma$. Let $D$ be the point on the line $AB$ such that $A$ is the midpoint of the segment $[DB]$ and $P$ is the point of intersection of $CD$ with $\Gamma$. Points $W$ and $L$ lie on the smaller arcs $\overarc{BC}$ and $\overarc{AB}$, respectively, and are such that $\overarc{BW} = \overarc{LA }= \overarc{AP}$. The $LC$ and $AW$ lines intersect at $Q$. Shows that $LQ = BQ$.
The circles $k_1$, $k_2$, $k_3$ are externally tangent: $k_1$ to $k_2$ at point $A$, $k_2$ to $k_3$ at point $B$, $k_3$ to $k_4$ at point $C$, $k_4$ to $k_1$ at point $D$. The lines $AB$ and $CD$ intersect at the point $S$. A line $ p $ is drawn through point $ S $, tangent to $ k_4 $ at point $ F $. Prove that $ |SE|=|SF| $.
Inside the equilateral triangle $ABC$, points $P$ and $Q$ are chosen so that the quadrilateral $APQC$ is convex, $AP = PQ = QC$ and $\angle PBQ = 30^o$. Prove that $AQ = BP$.
It is known that in the triangle $ABC$ the smallest side is $BC$. Let $X, Y, K$ and $L$ - points on the sides $AB, AC$ and on the rays $CB, BC$, respectively, are such that $BX = BK = BC =CY =CL$. The line $KX$ intersects the line $LY$ at the point $M$. Prove that the intersection point of the medians $\vartriangle KLM$ coincides with the center of the inscribed circle $\vartriangle ABC$.
Given a circle centered at $ O $ and two points $ A $ and $ B $ lying on it. $ A $ and $ B $ do not form a diameter. The point $ C $ is chosen on the circle so that the line $ AC $ divides the segment $ OB $ in half. Let lines $ AB $ and $ OC $ intersect at $ D $, and let lines $ BC $ and $ AO $ intersect at $ F $. Prove that $ AF = CD $.
$ABC$ is an acute triangle. Three semicircles are constructed outwardly on the sides $BC$, $CA$ and $AB$ respectively. Construct points $A'$ , $B'$ and $C' $ on these semicìrcles respectively so that $AB' = AC'$, $BC' = BA'$ and $CA'= CB'$.
Let $\vartriangle ABC$ be an isosceles triangle with $\angle BAC = 100^o$. Let $D, E$ be points on ray $\overrightarrow{AB}$ so that $BC = AD = BE$. Show that $BC \cdot DE = BD \cdot CE$
Altitudes $AA_1$ and $CC_1$ of an acute-angled triangle $ABC$ intersect at point $H$. A straight line passing through point $H$ parallel to line $A_1C_1$ intersects the circumscribed circles of triangles $AHC_1$ and $CHA_1$ at points $X$ and $Y$, respectively. Prove that points $X$ and $Y$ are equidistant from the midpoint of segment $BH$.
We consider the quadrilateral $ABCD$, with $AD = CD = CB$ and $AB \parallel CD$. Points $E$ and $F$ belong to the segments $CD$ and $CB$ so that angles $\angle ADE = \angle AEF$. Prove that:
a) $4CF \le CB$ ,
b) if $4CF = CB$, then $AE$ is the bisector of the angle $\angle DAF$.
In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$.
[img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]
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 $AX$.
Let $CD$ be an altitude of right-angled triangle $ABC$ with $\angle C = 90$. Regular triangles$ AED$ and $CFD$ are such that $E$ lies on the same side from $AB$ as $C$, and $F$ lies on the same side from $CD$ as $B$. The line $EF$ meets $AC$ at $L$. Prove that $FL = CL + LD$
Let $M$ be the midpoint of side BC of an acute-angled triangle $ABC$. Let $D$ and $E$ be the center of the excircle of triangle $AMB$ tangent to side $AB$ and the center of the excircle of triangle $AMC$ tangent to side $AC$, respectively. The circumscribed circle of triangle $ABD$ intersects line$ BC$ for the second time at point $F$, and the circumcircle of triangle $ACE$ is at point $G$. Prove that $| BF | = | CG|$.
Let $ABCD$ be the circumscribed quadrilateral with the incircle $(I)$. The circle $(I)$ touches $AB, BC, C D, DA$ at $M, N, P,Q$ respectively. Let $K$ and $L$ be the circumcenters of the triangles $AMN$ and $APQ$ respectively. The line $KL$ cuts the line $BD$ at $R$. The line $AI$ cuts the line $MQ$ at $J$. Prove that $RA = RJ$.
An arbitrary point $M$ inside an equilateral triangle $ABC$ was connected to vertices. Prove that on each side the triangle can be selected one point at a time so that the distances between them would be equal to $AM, BM, CM$.
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 $MD = KD$.
$ABCD$ is a parallelogram. A point $M$ is found on the side $AB$ or its extension such that $\angle MAD = \angle AMO$ where $O$ is the intersection point of the diagonals of the parallelogram. Prove that $MD = MG$.
(M Smurov)