Found problems: 1065
Let $ABC$ be an acute-angled triangle, and let $F$ be the foot of its altitude from the vertex $C$. Let $M$ be the midpoint of the segment $CA$. Assume that $CF=BM$. Then the angle $MBC$ is equal to angle $FCA$ if and only if the triangle $ABC$ is equilateral.
Given a convex quadrilateral $ABCD$, such that $OA = \frac{OB \cdot OD}{OC + CD}$ where $O$ is the intersection of the two diagonals. The circumcircle of triangle $ABC$ intersects $BD$ at point $Q$. Prove that $CQ$ bisects $\angle ACD$
Bisectors of angle $C$ and externalangle $A$ of trapezoid $ABCD$ with bases $BC$ and $AD$ intersect at point $M$, and the bisector of angle $B$ and external angle $D$ intersect at point $N$. Prove that the midpoint of the segment $MN$ is equidistant from the lines $AB$ and $CD$.
Let $ABCD$ be a cyclic quadrilateral with center $O$ such that $BD$ bisects $AC.$ Suppose that the angle bisector of $\angle ABC$ intersects the angle bisector of $\angle ADC$ at a single point $X$ different than $B$ and $D.$ Prove that the line passing through the circumcenters of triangles $XAC$ and $XBD$ bisects the segment $OX.$
[i]Proposed by Viktor Ahmeti and Leart Ajvazaj, Kosovo[/i]
In a triangle $ABC$ let $D$ be the intersection of the angle bisector of $\gamma$, angle at $C$, with the side $AB.$ And let $F$ be the area of the triangle $ABC.$ Prove the following inequality:
\[2 \cdot \ F \cdot \left( \frac{1}{AD} -\frac{1}{BD} \right) \leq AB.\]
Let $ ABC$ be a triangle, and $ I$ its incenter. Let the incircle of $ ABC$ touch side $ BC$ at $ D$, and let lines $ BI$ and $ CI$ meet the circle with diameter $ AI$ at points $ P$ and $ Q$, respectively. Given $ BI \equal{} 6, CI \equal{} 5, DI \equal{} 3$, determine the value of $ \left( DP / DQ \right)^2$.
In trapezoid $ABCD$, $AD$ is parallel to $BC$. Knowing that $AB=AD+BC$, prove that the bisector of $\angle A$ also bisects $CD$.
Let $H$ be an orhocenter of an acute triangle $ABC$ and $M$ midpoint of side $BC$. If $D$ and $E$ are foots of perpendicular of $H$ on internal and external angle bisector of angle $\angle BAC$, prove that $M$, $D$ and $E$ are collinear
Through the point $ K $ lying outside the circle $ \omega $, the tangents are drawn $ KB $ and $ KD $ to this circle ($ B $ and $ D $ are tangency points) and a line intersecting a circle at points $ A $ and $ C $. The bisector of angle $ ABC $ intersects the segment $ AC $ at the point $ E $ and circle $ \omega $ at $ F $. Prove that $ \angle FDE = 90^\circ $.
Let $ABC$ be an isosceles triangle with $AB=AC$. Consider a variable point $P$ on the extension of the segment $BC$ beyound $B$ (in other words, $P$ lies on the line $BC$ such that the point $B$ lies inside the segment $PC$). Let $r_{1}$ be the radius of the incircle of the triangle $APB$, and let $r_{2}$ be the radius of the $P$-excircle of the triangle $APC$. Prove that the sum $r_{1}+r_{2}$ of these two radii remains constant when the point $P$ varies.
[i]Remark.[/i] The $P$-excircle of the triangle $APC$ is defined as the circle which touches the side $AC$ and the [i]extensions[/i] of the sides $AP$ and $CP$.
Let $ABCD$ be a quadrilateral in which every angle is smaller than $180^o$. If the bisectors of angles $\angle DAB$ and $\angle DCB$ are parallel, prove that $\angle ADC = \angle ABC$
Let $ABCD$ be a parallelogram. Let $M$ be a point on the side $AB$ and $N$ be a point on the side $BC$ such that the segments $AM$ and $CN$ have equal lengths and are non-zero. The lines $AN$ and $CM$ meet at $Q$.
Prove that the line $DQ$ is the bisector of the angle $\measuredangle ADC$.
[i]Alternative formulation.[/i] Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$.
Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$.
Given is a triangle $ABC$ with its circumscribed circle and $| AC | <| AB |$. On the short arc $AC$, there is a variable point $D\ne A$. Let $E$ be the reflection of $A$ wrt the inner bisector of $\angle BDC$. Prove that the line $DE$ passes through a fixed point, regardless of point $D$.
Given $\triangle ABC$ with $AB < AC$, let $\omega$ be the circle centered at the midpoint $M$ of $BC$ with diameter $AC - AB$. The internal bisector of $\angle BAC$ intersects $\omega$ at distinct points $X$ and $Y$. Let $T$ be the point on the plane such that $TX$ and $TY$ are tangent to $\omega$. Prove that $AT$ is perpendicular to $BC$.
The perpendicular bisector to side $AC$ of triangle $ABC$ intersects side $BC$ at point $M$ (see fig.). The bisector of angle $\angle AMB$ intersects the circumcircle of triangle $ABC$ at point $K$. Prove that the line passing through the centers of the inscribed circles triangles $AKM$ and $BKM$, perpendicular to the bisector of angle $\angle AKB$.
[img]https://cdn.artofproblemsolving.com/attachments/b/4/b53ec7df0643a90b835f142d99c417a2a1dd45.png[/img]
In triangle $ABC$, $AD$ and $AH$ are the angle bisector and the altitude of vertex $A$, respectively. The perpendicular bisector of $AD$, intersects the semicircles with diameters $AB$ and $AC$ which are drawn outside triangle $ABC$ in $X$ and $Y$, respectively. Prove that the quadrilateral $XYDH$ is concyclic.
[i]Proposed by Mahan Malihi[/i]
The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD\equal{}AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$.
In a triangle $ABC$, points $H,M,L$ are the feet of the altitude from $C$, the median from $A$, and the angle bisector from $B$, respectively. Show that if triangle $HML$ is equilateral, then so is triangle $ABC$.
Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\omega$ with center $O$. The lines $AD$ and $BC$ meet at $E$, while the lines $AB$ and $CD$ meet at $F$. Let $P$ be a point on the segment $EF$ such that $OP \perp EF$. The circle $\Gamma_{1}$ passes through $A$ and $E$ and is tangent to $\omega$ at $A$, while $\Gamma_{2}$ passes through $C$ and $F$ and is tangent to $\omega$ at $C$. If $\Gamma_{1}$ and $\Gamma_{2}$ meet at $X$ and $Y$, prove that $PO$ is the bisector of $\angle XPY$.
[i]Proposed by Nikola Velov[/i]
Acute triangle $ABC$ satisfies $\angle A > \angle C$. Let $D, E, F$ be the points that the triangle's incircle intersects with $BC, CA, AB$, respectively, and $P$ some point on $AF$ different from $F$. The angle bisector of $\angle ABC$ meets $PQR$'s circumcircle $O$ at $L, R$. $L$ is the point closer to $B$ than $R$. $O$ meets $DF, DR$ at point $Q(\neq F, L), S(\neq R)$ respectively, and $PS$ hits segment $BC$ at $T$. Show that $T, Q, L$ are collinear.
Given a triangular pyramid $ABCD$. Sphere $S_1$ passing through points $A$, $B$, $C$, intersects edges $AD$, $BD$, $CD$ at points $K$, $L$, $M$, respectively; sphere $S_2$ passing through points $A$, $B$, $D$ intersects the edges $AC$, $BC$, $DC$ at points $P$, $Q$, $M$ respectively. It turned out that $KL \parallel PQ$. Prove that the bisectors of plane angles $KMQ$ and $LMP$ are the same.
In an acute triangle $ABC$ the angle $\angle C$ is greater than $\angle A$. Let $E$ be such that $AE$ is a diameter of the circumscribed circle $\Gamma$ of \vartriangle ABC. Let $K$ be the intersection of $AC$ and the tangent line at $B$ to $\Gamma$. Let $L$ be the orthogonal projection of $K$ on $AE$ and let $D$ be the intersection of $KL$ and $AB$. Prove that $CE$ is the bisector of $\angle BCD$.
Let $ABC$ be a triangle with $\angle BAC=117^\circ$. The angle bisector of $\angle ABC$ intersects side $AC$ at $D$. Suppose $\triangle ABD\sim\triangle ACB$. Compute the measure of $\angle ABC$, in degrees.
Let $F$ be an intersection point of altitude $CD$ and internal angle bisector $AE$ of right angled triangle $ABC$, $\angle ACB = 90^{\circ}$. Let $G$ be an intersection point of lines $ED$ and $BF$. Prove that area of quadrilateral $CEFG$ is equal to area of triangle $BDG$
Given triangle $ABC$. $D$ lies on $BC$ such that $AD$ bisects $BAC$. Given $AB=3$, $AC=9$, and
$BC=8$. Find $AD$.