Found problems: 670
7. The four Vertices of a quadrilateral [i]ABCD[/i] lie on the circle with diameter [i]AB[/i]. The diagonals of [i]ABCD[/i] intersect at [i]E[/i], and the lines [i]AD[/i] and [i]BC[/i] intersect at [i]F[/i]. Line [i]FE[/i] meets [i]AB[/i] at [i]K[/i] and line [i]DK[/i] meets the circle again at [i]L[/i].
Prove that [i]CL[/i] is perpendicular to [i]AB[/i].
The diagonals of a cyclic quadrilateral $ABCD$ intersect at $P$, and there exist a circle $\Gamma$ tangent to the extensions of $AB,BC,AD,DC$ at $X,Y,Z,T$ respectively. Circle $\Omega$ passes through points $A,B$, and is externally tangent to circle $\Gamma$ at $S$. Prove that $SP\perp ST$.
Let $A,B,C,D$ be points on a circle with radius $r$ in this order such that $|AB|=|BC|=|CD|=s$ and $|AD|=s+r$. Find all possible values of the interior angles of the quadrilateral $ABCD$.
Consider a cyclic quadrilateral $ABCD$ in which $BC = CD$ and $AB < AD$. Let $E$ be a point on the side $AD$ and $F$ a point on the line $BC$ such that $AE = AB = AF$. Prove that $EF \parallel BD$.
Given is a cyclic quadrilateral $ABCD$. The point $L_a$ lies in the interior of $BCD$ and is such that its distances to the sides of this triangle are proportional to the lengths of corresponding sides. The points $L_b, L_c$, and $L_d$ are defined analogously. Given that the quadrilateral $L_aL_bL_cL_d$ is cyclic, prove that the quadrilateral $ABCD$ has two parallel sides.
(N. Beluhov)
In the quadrilateral $ABCD$ the angles $B{}$ and $D{}$ are straight. The lines $AB{}$ and $DC{}$ intersect at $E$ and the lines $AD$ and $BC$ intersect at $F{}.$ The line passing through $B{}$ parallel to $C{}$D intersects the circumscribed circle $\omega$ of $ABF{}$ at $K{}$ and the segment $KE{}$ intersects $\omega$ at $P{}.$ Prove that the line $AP$ divides the segment $CE$ in half.
Let $ABCD$ be a convex cyclic quadrilateral with diagonals $AC$ and $BD$. Each of the four vertixes are reflected across the diagonal on which the do not lie.
(a) Investigate when the four points thus obtained lie on a straight line and give as simple an equivalent condition as possible to the cyclic quadrilateral $ABCD$ for it.
(b) Show that in all other cases the four points thus obtained lie on one circle.
(Theresia Eisenkölbl)
$ABCD$ is a cyclic quadrilateral, with diagonals $AC,BD$ perpendicular to each other. Let point $F$ be on side $BC$, the parallel line $EF$ to $AC$ intersect $AB$ at point $E$, line $FG$ parallel to $BD$ intersect $CD$ at $G$. Let the projection of $E$ onto $CD$ be $P$, projection of $F$ onto $DA$ be $Q$, projection of $G$ onto $AB$ be $R$. Prove that $QF$ bisects $\angle PQR$.
9.4, 10.3 Let $I$ be an incenter of $ABC$ ($AB<BC$), $M$ is a midpoint of $AC$, $N$ is a midpoint of circumcircle's arc $ABC$. Prove that $\angle IMA=\angle INB$.
([i]A. Badzyan[/i])
Let $ABCD$ be any convex quadrilateral and let $P, Q, R, S$ be points on the segments $AB, BC, CD$, and $DA$, respectively. It is given that the segments $PR$ and $QS$ dissect $ABCD$ into four quadrilaterals, each of which has perpendicular diagonals. Show that the points $P, Q, R, S$ are concyclic.
Two lines $m$ and $n$ intersect in a unique point $P$. A point $M$ moves along $m$ with constant speed, while another point $N$ moves along $n$ with the same speed. They both pass through the point $P$, but not at the same time.
Show that there is a fixed point $Q \ne P$ such that the points $P,Q,M$ and $N$ lie on a common circle all the time.
Let $ABCD$ be a cyclic quadrilateral with $\angle ABC = 60^o$ and $| BC | = | CD |$. Prove that $|CD| + |DA| = |AB|$
Let $ABCD$ be a cyclic quadrilateral. A circle passing through $A$ and $B$ meets $AC$ and $BD$ at points $E$ and $F$ respectively. The lines $AF$ and $BC$ meet at point $P$, and the lines $BE$ and $AD$ meet at point $Q$. Prove that $PQ$ is parallel to $CD$.
Quadrilateral $ABCD$ is inscribed in a circle. Let $\omega_A$, $\omega_B$, $\omega_C$, $\omega_D$ be the incircles of triangles $DAB$, $ABC$, $BCD$, $CDA$ respectively. The common external common tangent of $\omega_A$, $\omega_B$, different from line $AB$, meets the external common tangent of $\omega_A$, $\omega_D$, different from $AD$, at point $A'$. Similarly, the external common tangent of $\omega_B$, $\omega_C$ different from $BC$ meets the external common tangent of $\omega_C$, $\omega_D$ different from $CD$ at $C'$.
Prove that $AA'\parallel CC'$.
Given is a cyclic quadrilateral $ABCD$ with $|AB| = |BC|$. Point $E$ is on the arc $CD$ where $A$ and $B$ are not on. Let $P$ be the intersection point of $BE$ and $CD$ , let $Q$ be the intersection point of $AE$ and $BD$ . Prove that $PQ \parallel AC$.
A quadrilateral $ ABCD $ is inscribed in a circle. The lines $AB$ and $CD$ intersect at point $E$, and the lines $AD$ and $BC$ intersect at point $F$. The bisector of the angle $ AEC $ intersects the side $ BC $ at the point $ M $ and the side $ AD $ at the point $ N $; and the bisector of the angle $ BFD $ intersects the side $ AB $ at the point $ P $ and the side $ CD $ at the point $ Q $. Prove that the quadrilateral $MPNQ$ is a rhombus.
On sides $AB$ and $BC$ of a non-isosceles triangle $ABC$ are selected points $C_1$ and $A_1$ such that the quadrilateral $AC_1A_1C$ is cyclic. Lines $CC_1$ and $AA_1$ intersect at point $P$. Line $BP$ intersects the circumscribed circle of triangle $ABC$ at the point $Q$. Prove that the lines $QC_1$ and $CM$, where $M$ is the midpoint of $A_1C_1$, intersect at the circumscribed circles of triangle $ABC$.
The opposite sides of a quadrilateral inscribed in a circle intersect at points $K$ and $L$. Let $F$ be the midpoint of $KL$, $E$ and $G$ be the midpoints of the diagonals of the given quadrilateral. It is known that $FE = a$, $FG = b$. Calculate $KL$ in terms of $a$ and $b.$
(It is known that the points $F$, $E$ and $G$ lie on the same straight line. This is true for any quadrilateral, not necessarily inscribed. The indicated straight line is sometimes called the Newton−Gauss line. This fact can be used without proof in proving the problem, as it is known).
A half-circle of diameter $EF$ is placed on the side $BC$ of a triangle $ABC$ and it is tangent to the sides $AB$ and $AC$ in the points $Q$ and $P$ respectively. Prove that the intersection point $K$ between the lines $EP$ and $FQ$ lies on the altitude from $A$ of the triangle $ABC$.
[i]Albania[/i]
Let $M$ be the intersection of the diagonals of a cyclic quadrilateral $ABCD$. Find the length of $AD$, if it is known that $AB=2$ mm , $BC = 5$ mm, $AM = 4$ mm, and $\frac{CD}{CM}= 0.6$.
Given a quadrilateral $ABCD$, inscribed in a circle $\omega$ such that $AB=AD$ and $CB=CD$ . Take the point $P \in \omega$. Let the vertices of the quadrilateral $Q_1Q_2Q_3Q_4$ be symmetric to the point P wrt the lines $AB$, $BC$, $CD$, and $DA$, respectively.
a) Prove that the points symmetric to the point $P$ wrt lines $Q_1Q_22, Q_2Q_3, Q_3Q_4$ and $Q_4Q_1$, lie on one line.
b) Prove that when the point $P$ moves in a circle $\omega$, then all such lines pass through one common point.
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
Let $ABC$ be a given triangle with circumcenter $O$ and orthocenter $H$. Let $D, E$ and $F$ be the feet of the perpendiculars from $A, B$ and $C$ to the opposite sides, respectively. Let $A'$ be the reflection of $A$ with respect to $EF$. Prove that $HOA'D$ is a cyclic quadrilateral.
[i]Proposed by Imad Uddin Ahmad Hasin[/i]
Let $ABCD$ be a quadrilateral inscribed in a circle $\Omega.$ Let the tangent to $\Omega$ at $D$ meet rays $BA$ and $BC$ at $E$ and $F,$ respectively. A point $T$ is chosen inside $\triangle ABC$ so that $\overline{TE}\parallel\overline{CD}$ and $\overline{TF}\parallel\overline{AD}.$ Let $K\ne D$ be a point on segment $DF$ satisfying $TD=TK.$ Prove that lines $AC,DT,$ and $BK$ are concurrent.
Point $M$ on side $AB$ of quadrilateral $ABCD$ is such that quadrilaterals $AMCD$ and $BMDC$ are circumscribed around circles centered at $O_1$ and $O_2$ respectively. Line $O_1O_2$ cuts an isosceles triangle with vertex $M$ from angle $CMD$. prove that $ABCD$ is a cyclc quadrilateral.