Found problems: 1065
From a point $Q$ on a circle with diameter $AB$ different from $A$ and $B$, we draw a perpendicular to $AB$, $QH$, where $H$ lies on $AB$. The intersection points of the circle of diameter $AB$ and the circle of center $Q$ and radius $QH$ are $C$ and $D$. Prove that $CD$ bisects $QH$.
In a triangle $ABC$, the median $AD$ (with $D$ on $BC$) and the angle bisector $BE$ (with $E$ on $AC$) are perpedicular to each other. If $AD = 7$ and $BE = 9$, find the integer nearest to the area of triangle $ABC$.
Let $ABC$ be a triangle with circumcentre $O$. The points $D,E,F$ lie in the interiors of the sides $BC,CA,AB$ respectively, such that $DE$ is perpendicular to $CO$ and $DF$ is perpendicular to $BO$. (By interior we mean, for example, that the point $D$ lies on the line $BC$ and $D$ is between $B$ and $C$ on that line.)
Let $K$ be the circumcentre of triangle $AFE$. Prove that the lines $DK$ and $BC$ are perpendicular.
[i]Netherlands (Merlijn Staps)[/i]
Let $ABCD$ be a parallelogram such that $AB = 35$ and $BC = 28$. Suppose that $BD \perp BC$. Let $\ell_1$ be the reflection of $AC$ across the angle bisector of $\angle BAD$, and let $\ell_2$ be the line through $B$ perpendicular to $CD$. $\ell_1$ and $\ell_2$ intersect at a point $P$. If $PD$ can be expressed in simplest form as $\frac{m}{n}$, find $m + n$.
Let $D$ and $E$ be foots of altitudes from $A$ and $B$ of triangle $ABC$, $F$ be intersection point of angle bisector from $C$ with side $AB$, and $O$, $I$ and $H$ be circumcenter, center of inscribed circle and orthocenter of triangle $ABC$, respectively. If $\frac{CF}{AD}+ \frac{CF}{BE}=2$, prove that $OI = IH$.
On the side $AB$ of the non-isosceles triangle $ABC$, let the points $P$ and $Q$ be so that $AC = AP$ and $BC = BQ$. The perpendicular bisector of the segment $PQ$ intersects the angle bisector of the $\angle C$ at the point $R$ (inside the triangle). Prove that $\angle ACB + \angle PRQ = 180^o$.
In a triangle $ABC$, the angle bisector at $A,B,C$ meet the opposite sides at $A_1,B_1,C_1$, respectively. Prove that if the quadrilateral $BA_1B_1C_1$ is cyclic, then
$$\frac{AC}{AB+BC}=\frac{AB}{AC+CB}+\frac{BC}{BA+AC}.$$
Let $ABC$ be an acute triangle with $AB < AC$. The points $A_1$ and $A_2$ are located on the line $BC$ so that $AA_1$ and $AA_2$ are the inner and outer angle bisectors at $A$ for the triangle $ABC$. Let $A_3$ be the mirror image $A_2$ with respect to $C$, and let $Q$ be a point on $AA_1$ such that $\angle A_1QA_3 = 90^o$. Show that $QC // AB$.
Let $ABCD$ be a tangential quadrilateral with $BC> BA$. The point $P$ is on the segment $BC$, such that $BP = BA$ . Show that the bisector of $\angle BCD$, the perpendicular on line $BC$ through $P$ and the perpendicular on $BD$ through $A$, intersect at one point.
$ ABC$ is a triangle with $ \angle BAC \equal{} 10{}^\circ$, $ \angle ABC \equal{} 150{}^\circ$. Let $ X$ be a point on $ \left[AC\right]$ such that $ \left|AX\right| \equal{} \left|BC\right|$. Find $ \angle BXC$.
$\textbf{(A)}\ 15^\circ \qquad\textbf{(B)}\ 20^\circ \qquad\textbf{(C)}\ 25^\circ \qquad\textbf{(D)}\ 30^\circ \qquad\textbf{(E)}\ 35^\circ$
Let $ABC$ be a scalene triangle and $\omega$ is your incircle. The sides $BC,CA$ and $AB$ are tangents to $\omega$ in $X,Y,Z$ respectively. Let $M$ be the midpoint of $BC$ and $D$ is the intersection point of $BC$ with the angle bisector of $\angle BAC$. Prove that $\angle BAX=\angle MAC$ if and only if $YZ$ passes by the midpoint of $AD$.
Let $ABC$ be a triangle with $AB = AC$ ¸ $\angle BAC = 100^o$ and $AD, BE$ angle bisectors. Prove that $2AD <BE + EA$
The perpendicular line issued from the center of the circumcircle to the bisector of angle $C$ in a triangle $ABC$ divides the segment of the bisector inside $ABC$ into two segments with ratio of lengths $\lambda$. Given $b = AC$ and $a = BC$, find the length of side $c.$
Points $A$, $B$, $C$, and $D$ lie on a circle $\Gamma$, in that order, with $AB=5$ and $AD=3$. The angle bisector of $\angle ABC$ intersects $\Gamma$ at point $E$ on the opposite side of $\overleftrightarrow{CD}$ as $A$ and $B$. Assume that $\overline{BE}$ is a diameter of $\Gamma$ and $AC=AE$. Compute $DE$.
In $\vartriangle ABC, \angle A = 66^o$ and $| AB | <| AC |$. The outer bisector in $A$ intersects $BC$ in $D$ and $| BD | = | AB | + | AC |$. Determine the angles of $\vartriangle ABC$.
Let $ ABC$ be a triangle, and let the angle bisectors of its angles $ CAB$ and $ ABC$ meet the sides $ BC$ and $ CA$ at the points $ D$ and $ F$, respectively. The lines $ AD$ and $ BF$ meet the line through the point $ C$ parallel to $ AB$ at the points $ E$ and $ G$ respectively, and we have $ FG \equal{} DE$. Prove that $ CA \equal{} CB$.
[i]Original formulation:[/i]
Let $ ABC$ be a triangle and $ L$ the line through $ C$ parallel to the side $ AB.$ Let the internal bisector of the angle at $ A$ meet the side $ BC$ at $ D$ and the line $ L$ at $ E$ and let the internal bisector of the angle at $ B$ meet the side $ AC$ at $ F$ and the line $ L$ at $ G.$ If $ GF \equal{} DE,$ prove that $ AC \equal{} BC.$
Consider the triangle $ABC$, in which $AB > AC$. Let $P$ and $Q$ be the feet of the perpendiculars dropped from the vertices $B$ and $C$ on the bisector of the angle $BAC$, respectively. On the line $BC$ note point $B$ such that $AD \perp AP.$ Prove that the lines $BQ, PC$ and $AD$ intersect at one point.
Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD$. The two diagonals meet at $P$. Let $O$ be the foot of the perpendicular from $P$ to $EF$. Show that $\angle BOC=\angle AOD$.
Cyclic quadrilateral $ABCD$ has circumcircle $(O)$. Points $M$ and $N$ are the midpoints of $BC$ and $CD$, and $E$ and $F$ lie on $AB$ and $AD$ respectively such that $EF$ passes through $O$ and $EO=OF$. Let $EN$ meet $FM$ at $P$. Denote $S$ as the circumcenter of $\triangle PEF$. Line $PO$ intersects $AD$ and $BA$ at $Q$ and $R$ respectively. Suppose $OSPC$ is a parallelogram. Prove that $AQ=AR$.
Let $A$ and $B$ be points on the circle $k$ with center $O$, so that $AB> AO$. Let $C$ be the intersection of the bisectors of $\angle OAB$ and $k$, different from $A$. Let $D$ be the intersection of the straight line $AB$ with the circumcircle of the triangle $OBC$, different from $B$. Show that $AD = AO$ .
Let $\vartriangle ABC$ be a scalene triangle with $AB < BC < CA$. Let $D$ be the projection of $A$ onto the angle bisector of $\angle ABC$, and let $E$ be the projection of $A$ onto the angle bisector of $\angle ACB$. The line $DE$ cuts sides $AB$ and AC at $M$ and $N$, respectively. Prove that $$\frac{AB+AC}{BC} =\frac{DE}{MN} + 1$$
In an acute triangle $ABC$ , $AB>AC$ , $ \cos B+ \cos C=1$ , $E,F$ are on the extend line of $AB,AC$ such that $\angle ABF = \angle ACE = 90$ .
(1) Prove :$BE+CF=EF$ ;
(2) Assume the bisector of $\angle EBC$ meet $EF$ at $P$ , prove that $CP$ is the bisector of $\angle BCF$.
[img]https://cdn.artofproblemsolving.com/attachments/a/2/c554c2bc0b4e044c45f88138568f5234d544a8.png[/img]
Points $L$ and $H$ are marked on the sides $AB$ of an acute-angled triangle ABC so that $CL$ is a bisector and $CH$ is an altitude. Let $P,Q$ be the feet of the perpendiculars from $L$ to $AC$ and $BC$ respectively. Prove that $AP \cdot BH = BQ \cdot AH$.
I. Gorodnin
In triangle $ABC$, where $AB<AC$, let $X$, $Y$, $Z$ denote the points where the incircle is tangent to $BC$, $CA$, $AB$, respectively. On the circumcircle of $ABC$, let $U$ denote the midpoint of the arc $BC$ that contains the point $A$. The line $UX$ meets the circumcircle again at the point $K$. Let $T$ denote the point of intersection of $AK$ and $YZ$. Prove that $XT$ is perpendicular to $YZ$.
Let $ABC$ an acute triangle and $H$ its orthocenter. Let $E$ and $F$ be the intersection of lines $BH$ and $CH$ with $AC$ and $AB$ respectively, and let $D$ be the intersection of lines $EF$ and $BC$. Let $\Gamma_1$ be the circumcircle of $AEF$, and $\Gamma_2$ the circumcircle of $BHC$. The line $AD$ intersects $\Gamma_1$ at point $I \neq A$. Let $J$ be the feet of the internal bisector of $\angle{BHC}$ and $M$ the midpoint of the arc $\stackrel{\frown}{BC}$ from $\Gamma_2$ that contains the point $H$. The line $MJ$ intersects $\Gamma_2$ at point $N \neq M$. Show that the triangles $EIF$ and $CNB$ are similar.