Found problems: 1065
Let $ABC$ be a triangle with $AB=16$ and $AC=5$. Suppose that the bisectors of angle $\angle ABC$ and $\angle BCA$ meet at a point $P$ in the triangle's interior. Given that $AP=4$, compute $BC$.
Let $ABC$ be an isosceles triangle with $AB = AC$, $P$ be the midpoint of the minor arc $AB$ of its circumcircle, and $Q$ be the midpoint of $AC$. A circumcircle of triangle $APQ$ centered at $O$ meets $AB$ for the second
time at point $K$. Prove that lines $PO$ and $KQ$ meet on the bisector of angle $ABC$.
Let $\triangle ABC$ have $AB = 1$ cm, $BC = 2$ cm and $AC = \sqrt{3}$ cm. Points $D$, $E$ and $F$ lie on segments $AB$, $AC$ and $BC$ respectively are such that $AE = BD$ and $BF = AD$. The angle bisector of $\angle BAC$ intersects the circumcircle of $\triangle ADE$ for the second time at $M$ and the angle bisector of $\angle ABC$ intersects the circumcircle of $\triangle BDF$ at $N$. Determine the length of $MN$.
A circle centered at $ O$ has radius $ 1$ and contains the point $ A$. Segment $ AB$ is tangent to the circle at $ A$ and $ \angle{AOB} \equal{} \theta$. If point $ C$ lies on $ \overline{OA}$ and $ \overline{BC}$ bisects $ \angle{ABO}$, then $ OC \equal{}$
[asy]import olympiad;
unitsize(2cm);
defaultpen(fontsize(8pt)+linewidth(.8pt));
labelmargin=0.2;
dotfactor=3;
pair O=(0,0);
pair A=(1,0);
pair B=(1,1.5);
pair D=bisectorpoint(A,B,O);
pair C=extension(B,D,O,A);
draw(Circle(O,1));
draw(O--A--B--cycle);
draw(B--C);
label("$O$",O,SW);
dot(O);
label("$\theta$",(0.1,0.05),ENE);
dot(C);
label("$C$",C,S);
dot(A);
label("$A$",A,E);
dot(B);
label("$B$",B,E);[/asy]
$ \textbf{(A)}\ \sec^2\theta \minus{} \tan\theta \qquad \textbf{(B)}\ \frac {1}{2} \qquad \textbf{(C)}\ \frac {\cos^2\theta}{1 \plus{} \sin\theta} \qquad \textbf{(D)}\ \frac {1}{1 \plus{} \sin\theta} \qquad \textbf{(E)}\ \frac {\sin\theta}{\cos^2\theta}$
The bisector of the exterior angle at vertex $C$ of the triangle $ABC$ intersects the bisector of the interior angle at vertex $B$ in point $K$. Consider the diameter of the circumcircle of the triangle $BCK$ whose one endpoint is $K$. Prove that $A$ lies on this diameter.
In a convex quadrilateral $ABCD$ it is given that $\angle{CAB} = 40^{\circ}, \angle{CAD} = 30^{\circ}, \angle{DBA} = 75^{\circ}$, and $\angle{DBC}=25^{\circ}$. Find $\angle{BDC}$.
Two circles $S_1$ and $S_2$ with centers $O_1$ and $O_2$ respectively intersect at $A$ and $B$. The tangents at $A$ to $S_1$ and $S_2$ meet segments $BO_2$ and $BO_1$ at $K$ and $L$ respectively. Show that $KL \parallel O_1O_2.$
Let $L$ and $N$ be the mid-points of the diagonals $[AC]$ and $[BD]$ of the cyclic quadrilateral $ABCD$, respectively. If $BD$ is the bisector of the angle $ANC$, then prove that $AC$ is the bisector of the angle $BLD$.
In triangle $ABC$, the bisector $CL$ was drawn. The incircles of triangles $CAL$ and $CBL$ touch $AB$ at points $M$ and $N$ respectively. Points $M$ and $N$ are marked on the picture, and then the whole picture except the points $A, L, M$, and $N$ is erased. Restore the triangle using a compass and a ruler.
(V.Protasov)
Let ABC be a triangle. Let B' and C' denote the reflection of B and C in the internal angle bisector of angle A. Show that the triangles ABC and AB'C' have the same incenter.
In triangle $ABC$, $M$ is midpoint of $AC$, and $D$ is a point on $BC$ such that $DB=DM$. We know that $2BC^{2}-AC^{2}=AB.AC$. Prove that \[BD.DC=\frac{AC^{2}.AB}{2(AB+AC)}\]
Consider a triangle $ABC$ with $AB=AC$, and $D$ the foot of the altitude from the vertex $A$. The point $E$ lies on the side $AB$ such that $\angle ACE= \angle ECB=18^{\circ}$.
If $AD=3$, find the length of the segment $CE$.
In an isosceles, obtuse-angled triangle, the lengths of two internal angle bisectors are in a $2:1$ ratio. Find the obtuse angle of the triangle.
Let $ABC$ be an isosceles triangle with $AB = AC$. Let $X$ and $K$ points over $AC$ and $AB$, respectively, such that $KX = CX$. Bisector of $\angle AKX$ intersects line $BC$ at $Z$. Show that $XZ$ passes through the midpoint of $BK$.
Let be given a semicircle with the diameter $AB$, and points $C,D$ on it such that $AC = CD$. The tangent at $C$ intersects the line $BD$ at $E$. The line $AE$ intersects the arc of the semicircle at $F$. Prove that $CF < FD$.
Two circles $ \Omega_{1}$ and $ \Omega_{2}$ are externally tangent to each other at a point $ I$, and both of these circles are tangent to a third circle $ \Omega$ which encloses the two circles $ \Omega_{1}$ and $ \Omega_{2}$.
The common tangent to the two circles $ \Omega_{1}$ and $ \Omega_{2}$ at the point $ I$ meets the circle $ \Omega$ at a point $ A$. One common tangent to the circles $ \Omega_{1}$ and $ \Omega_{2}$ which doesn't pass through $ I$ meets the circle $ \Omega$ at the points $ B$ and $ C$ such that the points $ A$ and $ I$ lie on the same side of the line $ BC$.
Prove that the point $ I$ is the incenter of triangle $ ABC$.
[i]Alternative formulation.[/i] Two circles touch externally at a point $ I$. The two circles lie inside a large circle and both touch it. The chord $ BC$ of the large circle touches both smaller circles (not at $ I$). The common tangent to the two smaller circles at the point $ I$ meets the large circle at a point $ A$, where the points $ A$ and $ I$ are on the same side of the chord $ BC$. Show that the point $ I$ is the incenter of triangle $ ABC$.
Given triangle $ABC, BC$ is extended beyond $B$ to the point $D$ such that $BD = BA$. The bisectors of the exterior angles at vertices $B$ and $C$ intersect at the point $M$. Prove that quadrilateral $ADMC$ is cyclic. (4)
In obtuse triangle $ABC$, with the obtuse angle at $A$, let $D$, $E$, $F$ be the feet of the altitudes through $A$, $B$, $C$ respectively. $DE$ is parallel to $CF$, and $DF$ is parallel to the angle bisector of $\angle BAC$. Find the angles of the triangle.
In $ \triangle ABC$ we have $ AB \equal{} 7$, $ AC \equal{} 8$, and $ BC \equal{} 9$. Point $ D$ is on the circumscribed circle of the triangle so that $ \overline{AD}$ bisects $ \angle BAC$. What is the value of $ AD/CD$?
$ \textbf{(A)}\ \frac{9}{8}\qquad
\textbf{(B)}\ \frac{5}{3}\qquad
\textbf{(C)}\ 2\qquad
\textbf{(D)}\ \frac{17}{7}\qquad
\textbf{(E)}\ \frac{5}{2}$
Let $ABC$ be a scalene triangle, and $A_o$, $B_o,$ $C_o$ be the midpoints of $BC$, $CA$, $AB$ respectively. The bisector of angle $C$ meets $A_oCo$ and $B_oC_o$ at points $B_1$ and $A_1$ respectively. Prove that the lines $AB_1$, $BA_1$ and $A_oB_o$ concur.
The bisectors $AA_1$ and $CC_1$ of triangle $ABC$ intersect at point $I$. The circumscribed circles of triangles $AIC_1$ and $CIA_1$ intersect the arcs $AC$ and $BC$ (not containing points $B$ and $A$ respectively) of the circumscribed circle of triangle $ABC$ at points $C_2$ and $A_2$, respectively. Prove that lines $A_1A_2$ and $C_1C_2$ intersect on the circumscribed circle of triangle $ABC$.
Given a triangle $ABC$, the internal and external bisectors of angle $A$ intersect $BC$ at points $D$ and $E$ respectively. Let $F$ be the point (different from $A$) where line $AC$ intersects the circle $w$ with diameter $DE$. Finally, draw the tangent at $A$ to the circumcircle of triangle $ABF$, and let it hit $w$ at $A$ and $G$. Prove that $AF=AG$.
In a triangle $ABC$ it is given that $2AB=AC+BC$. Prove that the incentre of $\triangle ABC$, the circumcentre of $\triangle ABC$, and the midpoints of $AC$ and $BC$ are concyclic.
Let $ ABC$ be a triangle with circum-circle $ \Gamma$. Let $ M$ be a point in the interior of triangle $ ABC$ which is also on the bisector of $ \angle A$. Let $ AM, BM, CM$ meet $ \Gamma$ in $ A_{1}, B_{1}, C_{1}$ respectively. Suppose $ P$ is the point of intersection of $ A_{1}C_{1}$ with $ AB$; and $ Q$ is the point of intersection of $ A_{1}B_{1}$ with $ AC$. Prove that $ PQ$ is parallel to $ BC$.
On the plane there is an image of a circle with a marked center. Let an arbitrary angle be drawn on this plane. Using one ruler, construct the bisector of this angle.