Found problems: 3585
Given quadrilateral $ABCD$ such that $\angle BAD+2 \angle BCD=180 ^ \circ .$
Let $E$ be the intersection of $BD$ and the internal bisector of $\angle BAD$.
The perpendicular bisector of $AE$ intersects $CB,CD$ at $X,Y,$ respectively.
Prove that $A,C,X,Y$ are concyclic.
$ABCD$ is a cyclic quadrilateral inscribed in circle $O$. Let $O_1$ be the $A$-excenter of $ABC$ and $O_2$ the $A$-excenter of $ABD$. Show that $A, B, O_1, O_2$ is concyclic, and $O$ passes through the center of $(ABO_1O_2)$.
Recall that for concyclic $X, Y, Z, W$, the notation $(XYZW)$ denotes the circumcircle of $XYZW$.
We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.
[i]Proposed by Sammy Luo[/i]
$\omega$ is circumcirlce of triangle $ABC$. We draw a line parallel to $BC$ that intersects $AB,AC$ at $E,F$ and intersects $\omega$ at $U,V$. Assume that $M$ is midpoint of $BC$. Let $\omega'$ be circumcircle of $UMV$. We know that $R(ABC)=R(UMV)$. $ME$ and $\omega'$ intersect at $T$, and $FT$ intersects $\omega'$ at $S$. Prove that $EF$ is tangent to circumcircle of $MCS$.
A chord is drawn through the intersection point of the diagonals of an inscribed quadrilateral. It is known that the parts of this chord located outside the quadrilateral have lengths equal to $\frac13$ and $\frac14$ of this chord. In what ratio is this chord divided by the intersection point of the diagonals of the quadrilateral?
In triangle $ABC$, the angle bisectors are $BE$ and $CF$ (where $E, F$ are on the sides of the triangle), and their intersection point is $I$. Point $N$ lies on the circumcircle of $AEF$, and the angle $\angle IAN$ is right. The circumcircle of $AEF$ meets the line $NI$ a second time at the point $L$. Show that the circumcenter of $AIL$ lies on line $BC$.
Given a convex quadrilateral $ABCD$. We denote $I_A,I_B, I_C$ and $I_D$ centers of $\omega_A, \omega_B,\omega_C $and $\omega_D$,inscribed In the triangles $DAB, ABC, BCD$ and $CDA$, respectively.It turned out that $\angle BI_AA + \angle I_CI_AI_D = 180^\circ$. Prove that $\angle BI_BA + \angle I_CI_BI_D = 180^{\circ}$. (A. Kuznetsov)
Let $I$ be the incenter of a triangle $ABC$; $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ be the orthocenters of the triangles $BIC$, $AIC$, $AIB$; $M_{a}$, $M_{b}$, $M_{c}$ be the midpoints of $BC$, $CA$, $AB$, and $S_{a}$, $S_{b}$, $S_{c}$ be the midpoints of $AA^{\prime}$, $BB^{\prime}$, $CC^{\prime}$. Prove that $M_{a}S_{a}$, $M_{b}S_{b}$, $M_{c}S_{c}$ concur.
Proposed by: S Kuznetsov
The diagonals of bicentric quadrilateral $ABCD$ meet at point $L$. Given are three segments equal to $AL$, $BL$, $CL$. Restore the quadrilateral using a compass and a ruler.
Trapezoid $ABCD$ with bases $AB$ and $CD$ is inscribed in a circle. Prove that the quadrilateral formed by orthogonal projections of any point of this circle onto lines $AC, BC, AD$ and $BD$ is inscribed.
Let $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.
Let $ABC$ be an acute-angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $X$ and $Y$ be the points of intersection of the bisectors of the angles $\angle ACB$ and $\angle ABC$ with the line $DE$ and let $Z$ be the midpoint of $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $\angle A = 60^\circ$.
Let $\omega$ be the circumcircle of triangle $ABC$ with $AB<AC$. Let $I$ be its incenter and let $M$ be the midpoint of $BC$. The foot of the perpendicular from $M$ to $AI$ is $H$. The lines $MH, BI, AB$ form a triangle $T_b$ and the lines $MH, CI, AC$ form a triangle $T_c$. The circumcircle of $T_b$ meets $\omega$ at $B'$ and the circumcircle of $T_c$ meets $\omega$ at $C'$. Prove that $B', H, C'$ are collinear.
Let $ABC$ be a triangle and let $\Gamma$ be its circumcircle. Let $D$ be a point on $AB$ such that $CD$ is parallel to the line tangent to $\Gamma$ at $A$. Let $E$ be the intersection of $CD$ with $\Gamma$ distinct from $C$, and $F$ the intersection of $BC$ with the circumcircle of $\bigtriangleup ADC$ distinct from $C$. Finally, let $G$ be the intersection of the line $AB$ and the internal bisector of $\angle DCF$. Show that $E,\ G,\ F$ and $C$ lie on the same circle.
In an acute-angled and not isosceles triangle $ABC,$ we draw the median $AM$ and the height $AH.$
Points $Q$ and $P$ are marked on the lines $AB$ and $AC$, respectively, so that the $QM \perp AC$ and $PM \perp AB$.
The circumcircle of $PMQ$ intersects the line $BC$ for second time at point $X.$ Prove that $BH = CX.$
M. Didin
Let $A$, $B$ and $C$ be points lying on a circle $\Gamma$ with centre $O$. Assume that $\angle ABC > 90$. Let $D$ be the point of intersection of the line $AB$ with the line perpendicular to $AC$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $AO$. Let $E$ be the point of intersection of $l$ with the line $AC$, and let $F$ be the point of intersection of $\Gamma$ with $l$ that lies between $D$ and $E$.
Prove that the circumcircles of triangles $BFE$ and $CFD$ are tangent at $F$.
Let $ABC$ be a triangle and $O$ its circumcentre. A line tangent to the circumcircle of the triangle $BOC$ intersects sides $AB$ at $D$ and $AC$ at $E$. Let $A'$ be the image of $A$ under $DE$. Prove that the circumcircle of the triangle $A'DE$ is tangent to the circumcircle of triangle $ABC$.
A quick solution:
Let R be the foot of the perpend. from X to BC. Let's assume Q and R are in the interior of the segms AC and BC (respectively) and P in the ext of AD. P, R, Q are colinear (Simson's thm). PQ tangent to circle XRD iff XRQ=XDR iff Pi-XCA=XDR iff XBA=XDR=XDC=ADB iff XBC+ABC=ADB=DAC+ACB iff XAC+ABC=DAC+ACD iff ABC=ACD=ACB iff AB=AC. It's the same for all the other cases.
Pentagon $ABCDE$ is inscribed in circle $\Omega$. Line $AD$ meets $CE$ at $F$, and $P (\neq E, F)$ is a point on segment $EF$. The circumcircle of triangle $AFP$ meets $\Omega$ at $Q(\neq A)$ and $AC$ at $R(\neq A)$. Line $AD$ meets $BQ$ at $S$, and the circumcircle of triangle $DES$ meets line $BQ, BD$ at $T(\neq S), U(\neq D)$, respectively. Prove that if $F, P, T, S$ are concyclic, then $P, T, R, U$ are concyclic.
Let $G$ be the centroid of the triangle $ABC$ in which the angle at $C$ is obtuse and $AD$ and $CF$ be the medians from $A$ and $C$ respectively onto the sides $BC$ and $AB$. If the points $\ B,\ D, \ G$ and $\ F$ are concyclic, show that $\dfrac{AC}{BC} \geq \sqrt{2}$. If further $P$ is a point on the line $BG$ extended such that $AGCP$ is a parallelogram, show that triangle $ABC$ and $GAP$ are similar.
Let $ABC$ be a triangle with incenter $I$ and let $\Gamma$ be its circumcircle. Let $M$ be the midpoint of $BC$, $K$ the midpoint of the arc $BC$ which does not contain $A$, $L$ the midpoint of the arc $BC$ which contains $A$ and $J$ the reflection of $I$ by the line $KL$. The line $LJ$ intersects $\Gamma$ again at the point $T\neq L$. The line $TM$ intersects $\Gamma$ again at the point $S\neq T$. Prove that $S, I, M, K$ lie on the same circle.
In a triangle $ABC$ the incenter, the $B$-excenter and the $C$-excenter are $I, K$ and $L$, respectively. The perpendiculars at $B$ and $C$ to $BC$ intersect the lines $AC$ and $AB$ at $E$ and $F$, respectively. Prove that the circumcircles of $AEF, FIL, EIK$ concur.
Two points $K$ and $L$ are chosen inside triangle $ABC$ and a point $D$ is chosen on the side $AB$. Suppose that $B$, $K$, $L$, $C$ are concyclic, $\angle AKD = \angle BCK$ and $\angle ALD = \angle BCL$. Prove that $AK = AL$.
Trapezoid $ ABCD$, with $ \overline{AB}\parallel{}\overline{CD}$, is inscribed in circle $ \omega$ and point $ G$ lies inside triangle $ BCD$. Rays $ AG$ and $ BG$ meet $ \omega$ again at points $ P$ and $ Q$, respectively. Let the line through $ G$ parallel to $ \overline{AB}$ intersects $ \overline{BD}$ and $ \overline{BC}$ at points $ R$ and $ S$, respectively. Prove that quadrilateral $ PQRS$ is cyclic if and only if $ \overline{BG}$ bisects $ \angle CBD$.
The diagonals of the inscribed quadrangle $ABCD$ intersect at point $K$. Prove that the tangent at point $K$ to the circle circumscribed around the triangle $ABK$ is parallel to $CD$.
(A Zaslavsky)