Found problems: 1065
Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$. Let $M$ be a point on the side $BC$ such that $\angle BAM = \angle DAC$. Further, let $L$ be the second intersection point of the circumcircle of the triangle $CAM$ with the side $AB$, and let $K$ be the second intersection point of the circumcircle of the triangle $BAM$ with the side $AC$. Prove that $KL \parallel BC$.
The circle inscribed in the triangle $ABC$ is tangent to side $AC$ at point $B_1$, and to side $BC$ at point $A_1$. On the side $AB$ there is a point $K$ such that $AK = KB_1, BK = KA_1$. Prove that $ \angle ACB\ge 60$
From a point $P$ outside a circle, two tangents are drawn touching the circle at points $A$ and $C$. Let $B$ be a point on segment $AC$, and let segment $PB$ intersect the circle at point $Q$. The angle bisector of $\angle AQC$ intersects segment $AC$ at $R$. Show that $$\frac{AB}{BC} =\left(\frac{ AR}{RC}\right)^2$$
In a triangle $ABC$, let $AA', BB'$ and $CC'$ be the bisectors. Suppose $A'B' \cap CC' =P$ and $A'C' \cap BB'= Q$. Prove that $\angle PAC = \angle QAB$.
An arbitrary point $M$ is selected in the interior of the segment $AB$. The square $AMCD$ and $MBEF$ are constructed on the same side of $AB$, with segments $AM$ and $MB$ as their respective bases. The circles circumscribed about these squares, with centers $P$ and $Q$, intersect at $M$ and also at another point $N$. Let $N'$ denote the point of intersection of the straight lines $AF$ and $BC$.
a) Prove that $N$ and $N'$ coincide;
b) Prove that the straight lines $MN$ pass through a fixed point $S$ independent of the choice of $M$;
c) Find the locus of the midpoints of the segments $PQ$ as $M$ varies between $A$ and $B$.
Let $AD$ and $BE$ be angle bisectors in triangle $ABC$. Let $x$, $y$ and $z$ be distances from point $M$, which lies on segment $DE$, from sides $BC$, $CA$ and $AB$, respectively. Prove that $z=x+y$
The $A$-excircle of a triangle $ABC$ touches the side $BC$ at the point $K$ and the extended side $AB$ at the point $L$. The $B$-excircle touches the lines $BA$ and $BC$ at the points $M$ and $N$, respectively. The lines $KL$ and $MN$ meet at the point $X$.
Show that the line $CX$ bisects the angle $ACN$.
Two circles $ \left(\Omega_1\right),\left(\Omega_2\right) $ touch internally on the point $ T $. Let $ M,N $ be two points on the circle $ \left(\Omega_1\right) $ which are different from $ T $ and $ A,B,C,D $ be four points on $ \left(\Omega_2\right) $ such that the chords $ AB, CD $ pass through $ M,N $, respectively. Prove that if $ AC,BD,MN $ have a common point $ K $, then $ TK $ is the angle bisector of $ \angle MTN $.
* $ \left(\Omega_2\right) $ is bigger than $ \left(\Omega_1\right) $
We consider the quadrilateral $ABCD$, with $AD = CD = CB$ and $AB \parallel CD$. Points $E$ and $F$ belong to the segments $CD$ and $CB$ so that angles $\angle ADE = \angle AEF$. Prove that:
a) $4CF \le CB$ ,
b) if $4CF = CB$, then $AE$ is the bisector of the angle $\angle DAF$.
Let $ABCD$ be a convex quadrilateral, the incenters of $\triangle ABC$ and $\triangle ADC$ are $I,J$, respectively. It is known that $AC,BD,IJ$ concurrent at a point $P$. The line perpendicular to $BD$ through $P$ intersects with the outer angle bisector of $\angle BAD$ and the outer angle bisector $\angle BCD$ at $E,F$, respectively. Show that $PE=PF$.
Let $ABC$ be a triangle and $AD$ be the bisector of the triangle ($D \in (BC)$) Assume that $AB =14$ cm,
$AC = 35$ cm and $AD = 12$ cm; which of the following is the area of triangle $ABC$ in cm$^2$?
[b]A.[/b] $\frac{1176}{5}$ [b]B.[/b] $\frac{1167}{5}$ [b]C.[/b] $234$ [b]D.[/b] $\frac{1176}{7}$ [b]E.[/b] $236$
Triangle $ABC$ has $AB = 4$, $AC = 5$, and $BC = 6$. An angle bisector is drawn from angle $A$, and meets $BC$ at $M$. What is the nearest integer to $100 \frac{AM}{CM}$?
In triangle $\triangle ABC$ points $M,N$ lie on $BC$ st : $\angle BAM= \angle MAN= \angle NAC$ . Points $P,Q$ are on the angle bisector of $BAC$, on the same side of $BC$ as A , st :
$$\frac{1}{3} \angle BAC = \frac{1}{2} \angle BPC = \angle BQC$$
Let $E = AM \cap CQ$ and $F = AN \cap BQ$ . Prove that the common tangents to $(EPF), (EQF)$ and the circumcircle of $\triangle ABC$ , are concurrent.
In a triangle $ABC,$ incircle touches the sides $BC, CA, AB$ at $D, E, F,$ respectively. A circle $\omega$ passing through $A$ and tangent to line $BC$ at $D$ intersects the line segments $BF$ and $CE$ at $K$ and $L,$ respectively. The line passing through $E$ and parallel to $DL$ intersects the line passing through $F$ and parallel to $DK$ at $P.$ If $R_1, R_2, R_3, R_4$ denotes the circumradius of the triangles $AFD, AED, FPD, EPD,$ respectively, prove that $R_1R_4=R_2R_3.$
Restore the triangle with a compass and a ruler given the intersection point of altitudes and the feet of the median and angle bisectors drawn to one side. (No research required.)
In the triangle $ABC$ let $AD$ be the interior angle bisector of $\angle ACB$, where $D\in AB$. The circumcenter of the triangle $ABC$ coincides with the incenter of the triangle $BCD$. Prove that $AC^2 = AD\cdot AB$.
The incircle of triangle $ABC$ touches $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. The perpendicular from the incenter $I$ to the median from vertex $C$ meets the line $A_1B_1$ in point $K$. Prove that $CK$ is parallel to $AB$.
Let $ ABCD$ be a cyclic quadrilateral and $ O$ be the intersection of diagonal $ AC$ and $ BD$. The circumcircles of triangle $ ABO$ and the triangle $ CDO$ intersect at $ K$. Let $ L$ be a point such that the triangle $ BLC$ is similar to $ AKD$ (in that order). Prove that if $ BLCK$ is a convex quadrilateral, then it has an incircle.
Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $ABC$ with $AB > BC$. The angle bisector of $\angle ABC$ intersects $\Omega$ at $M \ne B$. Let $\Gamma$ be the circle with diameter $BM$. The angle bisectors of $\angle AOB$ and $\angle BOC$ intersect $\Gamma$ at points $P$ and $Q,$ respectively. The point $R$ is chosen on the line $P Q$ so that $BR = MR$. Prove that $BR\parallel AC$.
(Here we always assume that an angle bisector is a ray.)
[i]Proposed by Sergey Berlov, Russia[/i]
If $ AD$ is the altitude, $ BE$ the angle bisector, and $ CF$ the median of a triangle $ ABC$, prove that $ AD,BE,$ and $ CF$ are concurrent if and only if:
$ a^2(a\minus{}c)\equal{}(b^2\minus{}c^2)(a\plus{}c),$
where $ a,b,c$ are the lengths of the sides $ BC,CA,AB$, respectively.
Points $M,\ N,\ K,\ L$ are taken on the sides $AB,\ BC,\ CD,\ DA$ of a rhombus
$ABCD,$ respectively, in such a way that $MN\parallel LK$ and the distance between $MN$ and $KL$ is equal to the height of $ABCD.$ Show that the circumcircles of the triangles $ALM$ and $NCK$ intersect each other, while those of $LDK$ and $MBN$ do not.
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ , and let $O$ , $H$ be the triangle's circumcenter and orthocenter respectively . Let also $A^{'}$ be the point where the angle bisector of the angle $BAC$ meets $\omega$ . If $A^{'}H=AH$ , then find the measure of the angle $BAC$.
Let $ABC$ be a triangle such that $AB<AC$. The perpendicular bisector of the side $BC$ meets the side $AC$ at the point $D$, and the (interior) bisectrix of the angle $ADB$ meets the circumcircle $ABC$ at the point $E$. Prove that the (interior) bisectrix of the angle $AEB$ and the line through the incentres of the triangles $ADE$ and $BDE$ are perpendicular.
Let $ABC$ be a triangle, let $D$ be the touch point of the side $BC$ and the incircle of the triangle $ABC$, and let $J_b$ and $J_c$ be the incentres of the triangles $ABD$ and $ACD$, respectively. Prove that the circumcentre of the triangle $AJ_bJ_c$ lies on the bisector of the angle $BAC$.
Let $P$ be a point inside a triangle $ABC$ with $\angle C = 90^o$ such that $AP = AC$, and let $M$ be the midpoint of $AB$ and $CH$ be the altitude. Prove that $PM$ bisects $\angle BPH$ if and only if $\angle A = 60^o$.