Found problems: 1065
Let the two tangents from a point $A$ outside a circle $k$ touch $k$ at $M$ and $N$. A line $p$ through $A$ intersects $k$ at $B$ and $C$, and $D$ is the midpoint of $MN$. Prove that $MN$ bisects the angle $BDC$
$D$ is midpoint of $AC$ for $\triangle ABC$. Bisectors of $\angle ACB,\angle ABD$ are perpendicular. Find max value for $\angle BAC$
[i](S. Berlov)[/i]
On the extension of the angle bisector $AL$ of the triangle $ABC$, a point $P$ is placed such that $P L = AL$. Prove that the perimeter of triangle $PBC$ does not exceed the perimeter of triangle $ABC$.
In plane,let a circle $(O)$ and two fixed points $B,C$ lies in $(O)$
such that $BC$ not is the diameter.Consider a point $A$ varies in
$(O)$ such that $A\neq B,C$ and $AB\neq AC$.Call $D$ and $E$
respective is intersect of $BC$ and internal and external bisector
of $\widehat{BAC}$,$I$ is midpoint of $DE$.The line that pass through
orthocenter of $\triangle ABC$
and perpendicular with $AI$ intersects $AD,AE$ respective at $M,N$.
1/Prove that $MN$ pass through a fixed point
2/Determint the place of $A$ such that $S_{AMN}$ has maxium value
For a $\triangle ABC$ , let $A_1$ be the symmetric point of the intersection of angle bisector of $\angle BAC$ and $BC$ , where center of the symmetry is the midpoint of side $BC$, In the same way we define $B_1 $ ( on $AC$ ) and $C_1$ (on $AB$). Intersection of circumcircle of $\triangle A_1B_1C_1$ and line $AB$ is the set $\{Z,C_1 \}$, with $BC$ is the set $\{X,A_1\}$ and with $CA$ is the set $\{Y,B_1\}$. If the perpendicular lines from $X,Y,Z$ on $BC,CA$ and $ AB$ , respectively are concurrent , prove that $\triangle ABC$ is isosceles.
On the sides of a triangle $ABC$ right angled at $A$ three points $D, E$ and $F$ (respectively $BC, AC$
and $AB$) are chosen so that the quadrilateral $AFDE$ is a square. If $x$ is the length of the side of the square, show that
\[\frac{1}{x}=\frac{1}{AB}+\frac{1}{AC}\]
In triangle $ABC$ we have $AB = 2, AC = 6$ and $\angle A = 120^o$ . The bisector of angle $A$ intersects the side BC at the point $D$. Determine the length of $AD$. The answer must be given as a fraction with integer numerator and denominator.
Let $ABCD$ be an isosceles trapezoid with bases $AD> BC$. Let $X$ be the intersection of the bisectors of $\angle BAC$ and $BC$. Let $E$ be the intersection of$ DB$ with the parallel to the bisector of $\angle CBD$ through $X$ and let $F$ be the intersection of $DC$ with the parallel to the bisector of $\angle DCB$ through $X$. Show that quadrilateral $AEFD$ is cyclic.
Let $ABC$ be a triangle with $AB \ne AC$. Let the angle bisector of $\angle BAC$ intersects $BC$ at $P$ and intersects the perpendicular bisector of segment $BC$ at $Q$. Prove that $\frac{PQ}{AQ} =\left( \frac{BC}{AB + AC}\right)^2$
A rectangle $ABCD$ is given. Segment $DK$ is equal to $BD$ and lies on the half-line $DC$. $M$ is the midpoint of $BK$. Prove that $AM$ is the angle bisector of $\angle BAC$.
[i]Proposed by S. Berlov[/i]
The angle bisector of $A$ in triangle $ABC$ intersects $BC$ in the point $D$. The point $E$ lies on the side $AC$, and the lines $AD$ and $BE$ intersect in the point $F$. Furthermore, $\frac{|AF|}{|F D|}= 3$ and $\frac{|BF|}{|F E|}=\frac{5}{3}$. Prove that $|AB| = |AC|$.
[img]https://1.bp.blogspot.com/-evofDCeJWPY/XzT9dmxXzVI/AAAAAAAAMVY/ZN87X3Cg8iMiULwvMhgFrXbdd_f1f-JWwCLcBGAsYHQ/s0/2013%2BMohr%2Bp5.png[/img]
Angle bisectors $AA', BB'$and $CC'$ are drawn in triangle $ABC$ with angle $\angle B= 120^o$. Find $\angle A'B'C'$.
In the isosceles triangle $ABC$ ($AC = BC$), $AB =\sqrt3$ and the altitude $CD =\sqrt2$. Let $E$ and $F$ be the midpoints of $BC$ and $DB$, respectively and $G$ be the intersection of $AE$ and $CF$. Prove that $D$ belongs to the angle bisector of $\angle AGF$.
Let $ABCD$ be a rectangle of centre $O$, such that $\angle DAC=60^{\circ}$. The angle bisector of $\angle DAC$ meets $DC$ at $S$. Lines $OS$ and $AD$ meet at $L$, and lines $BL$ and $AC$ meet at $M$. Prove that lines $SM$ and $CL$ are parallel.
$\triangle ABC$ has $AB=4$ and $AC=6$. Let point $D$ be on line $AB$ so that $A$ is between $B$ and $D$. Let the angle bisector of $\angle BAC$ intersect line $BC$ at $E$, and let the angle bisector of $\angle DAC$ intersect line $BC$ at $F$. Given that $AE=AF$, find the square of the circumcircle's radius' length.
Let $D,E,F$ be the points of contact of the incircle of an acute-angled triangle $ABC$ with $BC,CA,AB$ respectively. Let $I_1,I_2,I_3$ be the incentres of the triangles $AFE, BDF, CED$, respectively. Prove that the lines $I_1D, I_2E, I_3F$ are concurrent.
Let $A$ and $B$ be two points on a line $\ell$, $M$ the midpoint of $AB$, and $X$ a point on segment $AB$ other than $M$. Let $\Omega$ be a semicircle with diameter $AB$. Consider a point $P$ on $\Omega$ and let $\Gamma$ be the circle through $P$ and $X$ that is tangent to $AB$. Let $Q$ be the second intersection point of $\Omega$ and $\Gamma$. The internal angle bisector of $\angle PXQ$ intersects $\Gamma$ at a point $R$. Let $Y$ be a point on $\ell$ such that $RY$ is perpendicular to $\ell$. Show that $MX > XY$
The angle bisector of vertex $A$ of $\triangle ABC$ cuts $[BC]$ at $D$. The circle passing through $A$ and touching to $BC$ at $D$ meets $[AB]$ and $[AC]$ at $P$ and $Q$, respectively. $AD$ and $PQ$ meet at $T$. If $|AB|=5, |BC|=6, |CA|=7$, then $\frac{|AT|}{|TD|}=?$
$ \textbf{(A)}\ \frac75 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ \frac72 \qquad \textbf{(E)}\ 4$
Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$. The internal angle bisectors of \(\angle DAB\), \(\angle ABC\), \(\angle BCD\), \(\angle CDA\) create a convex quadrilateral $Q_1$. The external bisectors of the same angles create another convex quadrilateral $Q_2$.
Prove $Q_1$, $Q_2$ are cyclic, and that $O$ is the midpoint of their circumcenters.
Let $ABCD$ be a cyclic quadrilatetral inscribed in a circle $k$. Let $M$ and $N$ be the midpoints of the arcs $AB$ and $CD$ which do not contain $C$ and $A$ respectively. If $MN$ meets side $AB$ at $P$, then show that
$\frac{AP}{BP}=\frac{AC+AD}{BC+BD}$
Let $ \triangle ABC$ be a triangle with incircle $ \omega(I,r)$and circumcircle $ \zeta(O,R)$.Let $ l_{a}$ be the angle bisector of $ \angle BAC$.Denote $ P\equal{}l_{a}\cap\zeta$.Let $ D$ be the point of tangency $ \omega$ with $ [BC]$.Denote $ Q\equal{}PD\cap\zeta$.Show that $ PI\equal{}QI$ if $ PD\equal{}r$.
Let $ABC$ be a triangle such that $AB = AC$. Let $D,E$ be points on $AB,AC$ respectively such that $DE = AC$. Let $DE$ meet the circumcircle of triangle $ABC$ at point $T$. Let $P$ be a point on $AT$. Prove that $PD + PE = AT$ if and only if $P$ lies on the circumcircle of triangle $ADE$.
In an acute triangle $ABC,$ let $D$ be a point on the side $BC.$ Let $M_1, M_2, M_3, M_4, M_5$ be the midpoints of the line segments $AD, AB, AC, BD, CD,$ respectively and $O_1, O_2, O_3, O_4$ be the circumcenters of triangles $ABD, ACD, M_1M_2M_4, M_1M_3M_5,$ respectively. If $S$ and $T$ are midpoints of the line segments $AO_1$ and $AO_2,$ respectively, prove that $SO_3O_4T$ is an isosceles trapezoid.
Triangle $ABC$ is inscribed in a circle with center $O'$. A circle with center $O$ is inscribed in triangle $ABC$. $AO$ is drawn, and extended to intersect the larger circle in $D$. Then, we must have:
$\text{(A)}\ CD=BD=O'D \qquad
\text{(B)}\ AO=CO=OD \qquad
\text{(C)}\ CD=CO=BD \qquad\\
\text{(D)}\ CD=OD=BD \qquad
\text{(E)}\ O'B=O'C=OD $
[asy]
size(200);
defaultpen(linewidth(0.8)+fontsize(12pt));
pair A=origin,B=(15,0),C=(5,9),O=incenter(A,B,C),Op=circumcenter(A,B,C);
path incirc = incircle(A,B,C),circumcirc = circumcircle(A,B,C),line=A--3*O;
pair D[]=intersectionpoints(circumcirc,line);
draw(A--B--C--A--D[0]^^incirc^^circumcirc);
dot(O^^Op,linewidth(4));
label("$A$",A,dir(185));
label("$B$",B,dir(355));
label("$C$",C,dir(95));
label("$D$",D[0],dir(O--D[0]));
label("$O$",O,NW);
label("$O'$",Op,E);[/asy]
In a non isosceles triangle $ ABC$ let $ w$ be the angle bisector of the exterior angle at $ C$. Let $ D$ be the point of intersection of $ w$ with the extension of $ AB$. Let $ k_A$ be the circumcircle of the triangle $ ADC$ and analogy $ k_B$ the circumcircle of the triangle $ BDC$. Let $ t_A$ be the tangent line to $ k_A$ in A and $ t_B$ the tangent line to $ k_B$ in B. Let $ P$ be the point of intersection of $ t_A$ and $ t_B$.
Given are the points $ A$ and $ B$. Determine the set of points $ P\equal{}P(C )$ over all points $ C$, so that $ ABC$ is a non isosceles, acute-angled triangle.