Found problems: 509
Let $ABCD$ be a convex quadrilateral inscibed in circle $(O)$ such that $DB = DA + DC$. The point $P$ lies on the ray $AC$ such that $AP = BC$. The point $E$ is on $(O)$ such that $BE \perp AD$. Prove that $DP$ is parallel to the angle bisector of $\angle BEC$.
The diagonals of quadrilateral $ABCD$ intersect in point $E$. Given that $|AB| =|CE|$, $|BE| = |AD|$, and $\angle AED = \angle BAD$, determine the ratio $|BC|:|AD|$.
Given a quadrilateral $ABCD$ inscribed in circle $\Gamma$.From a point P outside $\Gamma$, draw tangents $PA$ and $PB$ with $A$ and $B$ as touspoints. The line $PC$ intersects $\Gamma$ at point $D$. Draw a line through $B$ parallel to $PA$, this line intersects $AC$ and $AD$ at points $E$ and $F$ respectively. Prove that $BE = BF$.
$P$ and $Q$ are points on the longest side $AB$ of triangle $ABC$ such that $AQ = AC$ and $BP = BC$. Prove that the circumcentre of triangle $CPQ$ coincides with the incentre of triangle $ABC$.
Let $AD$ be the altitude $ABC$ of an acute triangle. On the line $AD$ are chosen different points $E$ and $F$ so that $|DE |= |DF|$ and point $E$ is in the interior of triangle $ABC$. The circumcircle of triangle $BEF$ intersects $BC$ and $BA$ for second time at points $K$ and $M$ respectively. The circumcircle of the triangle $CEF$ intersects the $CB$ and $CA$ for the second time at points $L$ and $N$ respectively. Prove that the lines $AD, KM$ and $LN$ intersect at one point.
Inside the triangle $ABC $, the point $P $ is selected so that $BC = AP $ and $\angle APC = 180 {} ^ \circ - \angle ABC $. On the side $AB $ there is a point $K $, for which $AK = KB + PC $. Prove that $\angle AKC = 90 {} ^ \circ $.
(Danilo Hilko)
In a quadrilateral $ABCD$, angle $B$ is equal to $150^o$, angle $C$ is right, and sides $AB$ and $CD$ are equal. Determine the angle between $BC$ and the line connecting the midpoints of sides $BC$ and $AD$.
Let $ABC$ be a triangle in which $\angle ABC = 45^o$ and $\angle BAC > 90^o$. Let $O$ be the midpoint of the side $[BC]$. Consider the point $M \in (AC)$ such that $\angle COM =\angle CAB$. Perpendicular from $M$ on $AC$ intersects line $AB$ at point $P$.
a) Find the measure of the angle $\angle BCP$.
b) Show that if $\angle BAC = 105^o$, then $PB = 2MO$.
Let $ABCD$ be a parallelogram. On side $AB$, point $M$ is taken so that $AD = DM$. On side $AD$ point $N$ is taken so that $AB = BN$. Prove that $CM = CN$.
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]
In the hexagon $ABCDEF$, all angles are equally large. The side lengths satisfy $AB = CD = EF = 3$ and $BC = DE = F A = 2$. The diagonals $AD$ and $CF$ intersect each other in the point $G$. The point $H$ lies on the side $CD$ so that $DH = 1$. Prove that triangle $EGH$ is equilateral.
In triangle $ABC$, point $I$ is the center of the inscribed circle. $AT$ is a segment tangent to the circle circumscribed around the triangle $BIC$ . On the ray $AB$ beyond the point$ B$ and on the ray $AC$ beyond the point $C$, we draw the segments $BD$ and $CE$, respectively, such that $BD = CE = AT$. Let the point $F$ be such that $ABFC$ is a parallelogram. Prove that points $D, E$ and $F$ lie on the same line.
(Dmitry Prokopenko)
On the bisector of the angle $ BAC $ of the triangle $ ABC $ we choose the points $ {{B} _ {1}}, \, \, {{C} _ {1}} $ for which $ B {{B} _ {1 }}\perp AB $, $ C {{C} _ {1}} \perp AC $. The point $ M $ is the midpoint of the segment $ {{B} _ {1}} {{C} _ {1}} $. Prove that $ MB = MC $.
Let $ABC$ be an acute triangle. Suppose the points $A',B',C'$ lie on its sides $BC,AC,AB$ respectively and the segments $AA',BB',CC'$ intersect in a common point $P$ inside the triangle. For each of those segments let us consider the circle such that the segment is its diameter, and the chord of this circle that contains the point $P$ and is perpendicular to this diameter. All three these chords occurred to have the same length. Prove that $P$ is the orthocenter of the triangle $ABC$.
(Grigory Galperin)
In the acute-angled triangle $ABC$, with $AB \ne AC$, $D$ is the foot of the angle bisector of angle $A$, and $E, F$ are the feet of the altitudes from $B$ and $C$, respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$. Prove that $ME = MF$.
Leonard Giugiuc
Points $A_1,A_2,A_3,A_4$ are not concyclic, the same for points $B_1,B_2,B_3,B_4$. For all $i, j, k$ the circumradii of triangles $A_iA_jA_k$ and $B_iB_jB_k$ are equal. Can we assert that $A_iA_j=B_iB_j$ for all $i, j$'?
A line passing through vertex $A$ of regular triangle $ABC$ doesn’t intersect segment $BC$. Points $M$ and $N$ lie on this line, and $AM = AN = AB$ (point $B$ lies inside angle $MAC$). Prove that the quadrilateral formed by lines $AB, AC, BN, CM$ is cyclic.
Given a hexagon $ABCDEF$, in which $AB = BC, CD = DE, EF = FA$, and angles $A$ and $C$ are right. Prove that lines $FD$ and $BE$ are perpendicular.
(B. Kukushkin)
Let $ABC$ be a triangle with $\angle BAC =60^\circ$. $A'$ is the symmetric point of $A$ wrt $\overline{BC}$. $D$ is the point in $\overline{AC}$ such that $\overline{AB}=\overline{AD}$. $H$ is the orthocenter of triangle $ABC$. $l$ is the external angle bisector of $\angle BAC$. $\{M\}=\overline{A'D}\cap l$,$\{N\}=\overline{CH} \cap l$. Show that $\overline{AM}=\overline{AN}$.
Let $ABC$ be a triangle with $AB > AC$. Let $D$ be a point on the side $AB$ such that $DB = DC$ and let $M$ be the midpoint of $AC$. The line parallel to $BC$ passing through $D$ intersects the line $BM$ in $K$. Show that $\angle KCD = \angle DAC.$
On the side $AB$ of the triangle $ABC$ mark the points $M$ and $N$, such that $BM = BC$ and $AN = AC$. Then on the sides $BC$ and $AC$ mark the points$ P$ and $Q$, respectively, such that $BP = BN$ and $AQ = AM$. Prove that the points $C, Q, M, N$ and $P$ lie on the same circle.
Consider $D$ the midpoint of the base $[BC]$ of the isosceles triangle ABC in which $\angle BAC < 90^o$. On the perpendicular from $B$ on the line $BC$ consider the point $E$ such that $\angle EAB= \angle BAC$, and on the line passing though $C$ parallel to the line $AB$ we consider the point $F$ such that $F$ and $D$ are on different side of the line $AC$ and $\angle FAC = \angle CAD$. Prove that $AE = CF$ and $BF = EF$
Let $ABC$ be a triangle with $AB=AC$. Let $D$ be a point on $BC$, and $E$ a point on $AD$ such that $\angle BED=\angle BAC=2\angle CED$. Prove that $BD=2CD$.
A quadrilateral $ABCD$ is circumscribed about a circle. Lines $AC$ and $DC$ meet at point $E$ and lines $DA$ and $BC$ meet at $F$, where $B$ is between $A$ and $E$ and between $C$ and $F$. Let $I_1, I_2$ and $I_3$ be the incenters of triangles $AFB, BEC$ and $ABC$, respectively. The line $I_1I_3$ intersects $EA$ at $K$ and $ED$ at $L$, whereas the line $I_2I_3$ intersects $FC$ at $M$ and $FD$ at $N$. Prove that $EK = EL$ if and only if $FM = FN$
In the triangle $ABC, \angle BAC$ is acute, the angle bisector of $\angle BAC$ meets $BC$ at $D, K$ is the foot of the perpendicular from $B$ to $AC$, and $\angle ADB = 45^o$. Point $P$ lies between $K$ and $C$ such that $\angle KDP = 30^o$. Point $Q$ lies on the ray $DP$ such that $DQ = DK$. The perpendicular at $P$ to $AC$ meets $KD$ at $L$. Prove that $PL^2 = DQ \cdot PQ$.