Found problems: 1065
In a triangle $ABC$, with $AB \ne BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \ne B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$.
Given triangle $ABC$ with $AB<AC$. The line passing through $B$ and parallel to $AC$ meets the external angle bisector of $\angle BAC$ at $D$. The line passing through $C$ and parallel to $AB$ meets this bisector at $E$. Point $F$ lies on the side $AC$ and satisfies the equality $FC=AB$. Prove that $DF=FE$.
Let $ABC$ be a triangle with interior angle bisectors $AA_1, BB_1, CC_1$ and incenter $I$. If $\sigma[IA_1B] + \sigma[IB_1C] + \sigma[IC_1A] = \frac{1}{2}\sigma[ABC]$, where $\sigma[ABC]$ denotes the area of $ABC$, show that $ABC$ is isosceles.
Points $D$ and $E$ are taken on sides $BC$ and $CA$ of a triangle $ BD\equal{}AE$. Segments $AD$ and $BE$ meet at $P$. The bisector of $\angle ACB$ intersects $AD$ and $BE$ at $Q$ and $R$ respectively. Prove that $ \frac{PQ}{PR}\equal{}\frac{AD}{BE}$.
A trapezoid $ABCD$ is given with $BC // AD$. Assume that the bisectors of the angles $BAD$ and $CDA$ intersect on the perpendicular bisector of the line segment $BC$. Prove that $|AB|= |CD|$ or $|AB| +|CD| =|AD|$.
The points $P$ and $Q$ lie on the sides $BC$ and $CD$ of the parallelogram $ABCD$ so that $BP = QD$. Show that the intersection point between the lines $BQ$ and $DP$ lies on the line bisecting $\angle BAD$.
$ M$ is the midpoint of base $ BC$ in a trapezoid $ ABCD$. A point $ P$ is chosen on the base $ AD$. The line $ PM$ meets the line $ CD$ at a point $ Q$ such that $ C$ lies between $ Q$ and $ D$. The perpendicular to the bases drawn through $ P$ meets the line $ BQ$ at $ K$. Prove that $ \angle QBC \equal{} \angle KDA$.
[i]Proposed by S. Berlov[/i]
In triangle ABC, the altitude, angle bisector and median from C divide the angle C into four equal angles. Find angle B.
$ABC$ is a triangle
$E, F$ are points in $AB$, such that $AE = EF = FB$
$D$ is a point at the line $BC$ such that $ED$ is perpendiculat to $BC$
$AD$ is perpendicular to $CF$.
The angle CFA is the triple of angle BDF. ($3\angle BDF = \angle CFA$)
Determine the ratio $\frac{DB}{DC}$.
%Edited!%
Let $ABC$ be a triangle where $M$ is the midpoint of $\overline{AC}$, and $\overline{CN}$ is the angle bisector of $\angle ACB$ with $N$ on $\overline{AB}$. Let $X$ be the intersection of the median $\overline{BM}$ and the bisector $\overline{CN}$. In addition $\bigtriangleup BXN$ is equilateral and $AC=2$. What is $BN^2$?
$\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad\textbf{(B)}\ \frac{2}{9} \qquad\textbf{(C)}\ \frac{5\sqrt{2} - 3\sqrt{3}}{8} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad\textbf{(E)}\ \frac{3\sqrt{3} - 4}{5}$.
In a rectangle $ABCD, M$ and $N$ are the midpoints of $AD$ and $BC$ respectively and $P$ is a point on line $CD$. The line $PM$ meets $AC$ at $Q$. Prove that MN bisects the angle $\angle QNP$.
$ABC$ is an acute-angled triangle. Let $D$ be the point on $BC$ such that $AD$ is the bisector of $\angle A$. Let $E, F$ be the feet of perpendiculars from $D$ to $AC,AB$ respectively. Suppose the lines $BE$ and $CF$ meet at $H$. The circumcircle of triangle $AFH$ meets $BE$ at $G$ (apart from $H$). Prove that the triangle constructed from $BG$, $GE$ and $BF$ is right-angled.
In triangle $ABC, M$ is the midpoint of $AB$ and $D$ the foot of the bisector of angle $\angle ABC$. If $MD$ and $BD$ are known to be perpendicular, calculate $\frac{AB}{BC}$.
Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.
Let $ABC$ be a right triangle with $A = 90^{\circ}$ and $D \in (AC)$. Denote by $E$ the reflection of $A$ in the line $BD$ and $F$ the intersection point of $CE$ with the perpendicular in $D$ to $BC$. Prove that $AF, DE$ and $BC$ are concurrent.
Points $A_1,~ B_1,~ C_1$ lie on the sides $BC,~ AC$ and $AB$ of a triangle $ABC$, respectively, such that $AB_1 -AC_1 = CA_1 -CB_1 = BC_1 -BA_1$. Let $I_A,~ I_B,~ I_C$ be the incenters of triangles $AB_1C_1,~ A_1BC_1$ and $A_1B_1C$ respectively. Prove that the circumcenter of triangle $I_AI_BI_C$, is the incenter of triangle $ABC$.
Let $ABC$ be an acute-angled triangle in which $\angle ABC$ is the largest angle. Let $O$ be its circumcentre. The perpendicular bisectors of $BC$ and $AB$ meet $AC$ at $X$ and $Y$ respectively. The internal angle bisectors of $\angle AXB$ and $\angle BYC$ meet $AB$ and $BC$ at $D$ and $E$ respectively. Prove that $BO$ is perpendicular to $AC$ if $DE$ is parallel to $AC$.
In the figure on the right, $O$ is the center of the circle, $OK$ and $OA$ are perpendicular to one another, $M$ is the midpoint of $OK$, $BN$ is parallel to $OK$, and $\angle AMN=\angle NMO$. Determine the measure of $\angle A B N$ in degrees.
[asy]
defaultpen(linewidth(0.7)+fontsize(10));
pair O=origin, A=dir(90), K=dir(180), M=0.5*dir(180), N=2/5*dir(90), B=dir(degrees((2/5, sqrt(21/25)))+90);
draw(K--O--A--M--N--B--A^^Circle(origin,1));
label("$A$", A, dir(O--A));
label("$K$", K, dir(O--K));
label("$B$", B, dir(O--B));
label("$N$", N, E);
label("$M$", M, S);
label("$O$", O, SE);[/asy]
Suppose $ABCD$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\ell_1$ and $\ell_2$, which are also $a$ units apart. The square $ABCD$ is placed on the plane so that sides $AB$ and $AD$ intersect $\ell_1$ at $E$ and $F$ respectively. Also, sides $CB$ and $CD$ intersect $\ell_2$ at $G$ and $H$ respectively. Let the perimeters of $\triangle AEF$ and $\triangle CGH$ be $m_1$ and $m_2$ respectively.
Prove that no matter how the square was placed, $m_1+m_2$ remains constant.
$ABCD$ is a quadrilateral. The bisectors of $\angle A$ and $\angle B$ meet at $E$. The line through $E$ parallel to $CD$ meets $AD$ at $L$ and $BC$ at $M$. Show that $LM = AL + BM$.
In $\triangle ABC$ we have that $CC_{1}$ is an angle bisector. The points $P\in C_{1}B$, $Q\in BC$, $R\in AC$, $S\in AC_{1}$ satisfy $C_{1}P=PQ=QC$ and $CR=RS=SC_{1}$. Prove that $CC_{1}$ bisects $\angle SCP$.
Let $\vartriangle ABC$ be an acute triangle, and let $P$ be the foot of altitude from $C$ to $AB$. Let $\omega$ be the circle with diameter $BC$. The tangents from $A$ to $\omega$ are drawn touching $\omega$ at $D$ and $E$. Lines $AD$ and $AE$ intersect line $BC$ at $M$ and $N$ respectively, so that $B$ lies between $M$ and $C$. Let $CP$ intersect $DE$ at $Q, ME$ intersect $ND$ at $R$, and let $QR$ intersect $BC$ at $S$. Show that $QS$ bisects $\angle DSE$
In a triangle $ABC$, points $X$ and $Y$ are on $BC$ and $CA$ respectively such that $CX=CY$,$AX$ is not perpendicular to $BC$ and $BY$ is not perpendicular to $CA$.Let $\Gamma$ be the circle with $C$ as centre and $CX$ as its radius.Find the angles of triangle $ABC$ given that the orthocentres of triangles $AXB$ and $AYB$ lie on $\Gamma$.
In a convex quadrilateral $ABCD$ the angles $\angle A$ and $\angle C$ are equal and the bisector of $\angle B$ passes through the midpoint of the side $CD$. If it is known that $CD = 3AD$, calculate $\frac{AB}{BC}$.
Let $\triangle ABC$ be a triangle and $\omega$ its circumcircle. The exterior angle bisector of $\angle BAC$ intersects $\omega$ at point $D$. Let $X$ be the foot of the altitude from $C$ to $AD$ and let $F$ be the intersection of the internal angle bisector of $\angle BAC$ and $BC$. Show that $BX$ bisects segment $AF$.