Found problems: 1065
Let $ ABC$ be an isosceles right triangle and $M$ be the midpoint of its hypotenuse $AB$. Points $D$ and $E$ are taken on the legs $AC$ and $BC$ respectively such that $AD=2DC$ and $BE=2EC$. Lines $AE$ and $DM$ intersect at $F$. Show that $FC$ bisects the $\angle DFE$.
Given triangle $ABC$ with $|AC| > |BC|$. The point $M$ lies on the angle bisector of angle $C$, and $BM$ is perpendicular to the angle bisector. Prove that the area of triangle AMC is half of the area of triangle $ABC$.
[img]https://cdn.artofproblemsolving.com/attachments/4/2/1b541b76ec4a9c052b8866acbfea9a0ce04b56.png[/img]
$A, B, C$ are adjacent vertices of a regular $2n$-gon and $D$ is the vertex opposite to $B$ (so that $BD$ passes through the center of the $2n$-gon). $X$ is a point on the side $AB$ and $Y$ is a point on the side $BC$ so that $XDY = \frac{\pi}{2n}$. Show that $DY$ bisects $\angle XYC$.
Let $O$ be a point inside an acute angle with the vertex $A$ and $H, N$ be the feet of the perpendiculars drawn from $O$ onto the sides of the angle. Let point $B$ belong to the bisector of the angle, $K$ be the foot of the perpendicular from $B$ onto either side of the angle. Denote by $P,F$ the midpoints of the segments $AK,HN$ respectively. Known that $ON + OH = BK$, prove that $PF$ is perpendicular to $AB$.
Ya. Konstantinovski
In the triangle $ABC$, $\angle .ACB = 60^o$, and the bisectors $AA_1$ and $BB_1$ intersect at the point $M$. Prove that $MB_1 = MA_1$.
In a triangle $ABC$, the angle bisector of $A$ and the cevians $BD$ and $CE$ concur at a point $P$ inside the triangle. Show that the quadrilateral $ADPE$ has an incircle if and only if $AB=AC$.
(F.Nilov, A.Zaslavsky) Let $ CC_0$ be a median of triangle $ ABC$; the perpendicular bisectors to $ AC$ and $ BC$ intersect $ CC_0$ in points $ A'$, $ B'$; $ C_1$ is the meet of lines $ AA'$ and $ BB'$. Prove that $ \angle C_1CA \equal{} \angle C_0CB$.
A right-angled triangle has an angle equal to $30^\circ.$ Prove that one of the bisectors of the triangle is twice as short as another one.
[i]Egor Bakaev[/i]
Let $ABCD$ be a convex quadrilateral and $K$ be the common point of rays $AB$ and $DC$. There exists a point $P$ on the bisectrix of angle $AKD$ such that lines $BP$ and $CP$ bisect segments $AC$ and $BD$ respectively. Prove that $AB = CD$.
Let $\triangle ABC$ have incenter $I$ and centroid $G$. Suppose that $P_A$ is the foot of the perpendicular from $C$ to the exterior angle bisector of $B$, and $Q_A$ is the foot of the perpendicular from $B$ to the exterior angle bisector of $C$. Define $P_B$, $P_C$, $Q_B$, and $Q_C$ similarly. Show that $P_A, P_B, P_C, Q_A, Q_B,$ and $Q_C$ lie on a circle whose center is on line $IG$.
$I$ - incenter , $M$- midpoint of arc $BAC$ of circumcircle, $AL$ - angle bisector of triangle $ABC$. $MI$ intersect circumcircle in $K$. Circumcircle of $AKL$ intersect $BC$ at $L$ and $P$.
Prove that $\angle AIP=90$
In triangle $ABC$ we havev $\angle A=\frac{\pi}{3}$. Construct $E$ and $F$ on continue of $AB$ and $AC$ respectively such that $BE=CF=BC$. Suppose that $EF$ meets circumcircle of $\triangle ACE$ in $K$. ($K\not \equiv E$). Prove that $K$ is on the bisector of $\angle A$.
In triangle $\triangle ABC$ , $I$ is the incenter and $M$ is the midpoint of arc $(BC)$ in the circumcircle of $(ABC)$not containing $A$. Let $X$ be an arbitrary point on the external angle bisector of $A$. Let $BX \cap (BIC) = T$. $Y$ lies on $(AXC)$ , different from $A$ , st $MA=MY$ . Prove that $TC || AY$
(Assume that $X$ is not on $(ABC)$ or $BC$)
Excircle of $ABC$ is tangent to the side $BC$ at point $K$ and is tangent to the extension of $AB$ at point $L$. Another excircle is tangent to extensions of sides $AB$ and $BC$ at points $M$ and $N$. Lines $KL$ and $MN$ intersect at point $X$. Prove that $CX$ is the bisector of angle $ACN$.
[I]Proposed by S. Berlov[/i]
Consider a triangle $\triangle ABC$, let the incircle centered at $I$ touch the sides $BC,CA,AB$ at points $D,E,F$ respectively. Let the angle bisector of $\angle BIC$ meet $BC$ at $M$, and the angle bisector of $\angle EDF$ meet $EF$ at $N$. Prove that $A,M,N$ are collinear.
In an acute triangle $ABC$ the angle $C$ is greater than the angle $A$. Let $AE$ be a diameter of the circumcircle of the triangle. Let the intersection point of the ray $AC$ and the tangent of the circumcircle through the vertex $B$ be $K$. The perpendicular to $AE$ through $K$ intersects the circumcircle of the triangle $BCK$ for the second time at point $D$. Prove that $CE$ bisects the angle $BCD$.
Consider a line segment $[AB]$ with midpoint $M$ and perpendicular bisector $m$. For each point$ X \ne M$ on m consider we are the intersection point $Y$ of the line $BX$ with the bisector from the angle $\angle BAX$. As $X$ approaches $M$, then approaches $Y$ to a point of $[AB]$. Which?
[img]https://cdn.artofproblemsolving.com/attachments/a/3/17d72a23011a9ec22deb20184717cc9c020a2b.png[/img]
[hide=original wording]Beschouw een lijnstuk [AB] met midden M en middelloodlijn m. Voor elk punt X 6= M op m beschouwenwe het snijpunt Y van de rechte BX met de bissectrice van de hoek < BAX . Als X tot M nadert, dan nadert
Y tot een punt van [AB]. Welk? [/hide]
A circle is inscribed in quadrilateral $ABCD,$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q.$ Given that $AP=19, PB=26, CQ=37,$ and $QD=23,$ find the square of the radius of the circle.
$\vartriangle ABC$ has side lengths $AB = 4$ and $AC = 9$. Angle bisector $AD$ bisects angle $A$ and intersects $BC$ at $D$. Let $k$ be the ratio $\frac{BD}{AB}$ . Given that the length $AD$ is an integer, find the sum of all possible $k^2$
.
In the triangle $ABC$ , the angle bisector $AD$ divides the side $BC$ into the ratio $BD: DC = 2: 1$. In what ratio, does the median $CE$ divide this bisector?
Let $\vartriangle ABC$ be a triangle such that angle $\angle ABC$ is less than angle $\angle ACB$. The bisector of angle $\angle BAC$ cuts side $BC$ at $D$. Let $E$ be on side $AB$ such that $\angle EDB = 90^o$ and $F$ on side $AC$ such that $\angle BED = \angle DEF$. Prove that $\angle BAD = \angle FDC$.
In a triangle $ABC$, let $D$ be the midpoint of the side $BC$, and let $E$ be a point on the side $AC$. The lines $BE$ and $AD$ meet at a point $F$.
Prove: If $\frac{BF}{FE}=\frac{BC}{AB}+1$, then the line $BE$ bisects the angle $ABC$.
P and Q are 2 points in the area bounded by 2 rays, e and f, coming out from a point O. Describe how to construct, with a ruler and a compass only, an isosceles triangle ABC, such that his base AB is on the ray e, the point C is on the ray f, P is on AC, and Q on BC.
$AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC+AD-AB-AC = 0$, find the range of $\angle BAC$.
[i]Alternative formulation.[/i] Let $AD$ be the altitude of triangle $ABC$ to the side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.
Vertex $B$ of the $\angle ABC$ lies out the circle, and the $[BA)$ and $[BC)$ beams intersect it. Point $K$ belongs to the intersection of the $[BA)$ beam and the circumference. Chord $KP$ is orthogonal to the angle bisector of $\angle ABC$ . Line $(KP)$ intersects the beam $BC$ in the point $M$. Prove that the segment $[PM]$ is twice as long as the distance from the circle centre to the angle bisector of $\angle ABC$ .