In triangle $ABC$, $K$ and $L$ are points on the side $BC$ ($K$ being closer to $B$ than $L$) such that $BC \cdot KL = BK \cdot CL$ and $AL$ bisects $\angle KAC$. Show that $AL \perp AB.$

An angle is given in a plane. Using only a compass, one must find out $(a)$ if this angle is acute. Find the minimal number of circles one must draw to be sure. $(b)$ if this angle equals $31^{\circ}$.(One may draw as many circles as one needs).

Let there be a convex quadrilateral $ABCD$. The angle bisector of $\angle A$ meets the angle bisector of $\angle B$, the angle bisector of $\angle D$ at $P, Q$ respectively. The angle bisector of $\angle C$ meets the angle bisector of $\angle D$, the angle bisector of $\angle B$ at $R, S$ respectively. $P, Q, R, S$ are all distinct points. $PR$ and $QS$ meets perpendicularly at point $Z$. Denote $l_A, l_B, l_C, l_D$ as the exterior angle bisectors of $\angle A, \angle B, \angle C, \angle D$. Denote $E = l_A \cap l_B$, $F= l_B \cap l_C$, $G = l_C \cap l_D$, and $H= l_D \cap l_A$. Let $K, L, M, N$ be the midpoints of $FG, GH, HE, EF$ respectively. Prove that the area of quadrilateral $KLMN$ is equal to $ZM \cdot ZK + ZL \cdot ZN$.

In a triangle $\triangle ABC$ with orthocenter $H$, let $BH$ and $CH$ intersect $AC$ and $AB$ at $E$ and $F$, respectively. If the tangent line to the circumcircle of $\triangle ABC$ passing through $A$ intersects $BC$ at $P$, $M$ is the midpoint of $AH$, and $EF$ intersects $BC$ at $G$, then prove that $PM$ is parallel to $GH$. [i]Proposed by Sreejato Bhattacharya[/i]

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 $IG$.

In the figure $ \overline{AB} \equal{} \overline{AC}$, angle $ BAD \equal{} 30^{\circ}$, and $ \overline{AE} \equal{} \overline{AD}$. [asy]unitsize(20); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair A=(3,3),B=(0,0),C=(6,0),D=(2,0),E=(5,1); draw(A--B--C--cycle); draw(A--D--E); label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",D,S); label("$E$",E,NE);[/asy]Then angle $ CDE$ equals: $ \textbf{(A)}\ 7\frac {1}{2}^{\circ} \qquad\textbf{(B)}\ 10^{\circ} \qquad\textbf{(C)}\ 12\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 15^{\circ} \qquad\textbf{(E)}\ 20^{\circ}$

Let $S$ be the sum of the interior angles of a polygon $P$ for which each interior angle is $7\frac{1}{2}$ times the exterior angle at the same vertex. Then $ \textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad$ $\textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad$ $\textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad$ $\textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad$ $\textbf{(E)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may or may not be regular} $

Let $ k$ be the inscribed circle of non-isosceles triangle $ \triangle ABC$, which center is $ S$. Circle $ k$ touches sides $ BC,CA,AB$ in points $ P,Q,R$ respectively. Line $ QR$ intersects $ BC$ in point $ M$. Let a circle which contains points $ B$ and $ C$ touch $ k$ in point $ N$. Circumscribed circle of $ \triangle MNP$ intersects line $ AP$ in point $ L$, different from $ P$. Prove that points $ S,L$ and $ M$ are collinear.

Let $\omega_{1}$ be a circle with centre $O$. $P$ is a point on $\omega_{1}$. $\omega_{2}$ is a circle with centre $P$, with radius smaller than $\omega_{1}$. $\omega_{1}$ meets $\omega_{2}$ at points $T$ and $Q$. Let $TR$ be a diameter of $\omega_{2}$. Draw another two circles with $RQ$ as the radius, $R$ and $P$ as the centres. These two circles meet at point $M$, with $M$ and $Q$ lie on the same side of $PR$. A circle with centre $M$ and radius $MR$ intersects $\omega_{2}$ at $R$ and $N$. Prove that a circle with centre $T$ and radius $TN$ passes through $O$.

Let $ABCD$ be a convex quadrilateral, $I_1$ and $I_2$ be the incenters of triangles $ABC$ and $DBC$ respectively. The line $I_1I_2$ intersects the lines $AB$ and $DC$ at points $E$ and $F$ respectively. Given that $AB$ and $CD$ intersect in $P$, and $PE=PF$, prove that the points $A$, $B$, $C$, $D$ lie on a circle.

Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$. [i]Proposed by Vyacheslev Yasinskiy, Ukraine[/i]

A circle which is tangent to sides $AB$ and $BC$ of triangle $ABC$ is also tangent to its circumcircle at point $T$. If $I$ in the incenter of triangle $ABC$, show that $\angle ATI=\angle CTI$.

In triangle $ABC$ the ratio $AC:CB$ is $3:4$. The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ ($A$ is between $P$ and $B$). The ratio $PA:AB$ is: ${{ \textbf{(A)}\ 1:3 \qquad\textbf{(B)}\ 3:4 \qquad\textbf{(C)}\ 4:3 \qquad\textbf{(D)}\ 3:1 }\qquad\textbf{(E)}\ 7:1 } $

Cirlce $\Omega$ is inscribed in triangle $ABC$ with $\angle BAC=40$. Point $D$ is inside the angle $BAC$ and is the intersection of exterior bisectors of angles $B$ and $C$ with the common side $BC$. Tangent form $D$ touches $\Omega$ in $E$. FInd $\angle BEC$.

Triangle $ ABC$ has $ \angle C \equal{} 60^{\circ}$ and $ BC \equal{} 4$. Point $ D$ is the midpoint of $ BC$. What is the largest possible value of $ \tan{\angle BAD}$? $ \textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} \minus{} 3} \qquad \textbf{(E)}\ 1$

Let $ABCD$ be a circumscribed quadrilateral (i. e. a quadrilateral which has an incircle). The exterior angle bisectors of the angles $DAB$ and $ABC$ intersect each other at $K$; the exterior angle bisectors of the angles $ABC$ and $BCD$ intersect each other at $L$; the exterior angle bisectors of the angles $BCD$ and $CDA$ intersect each other at $M$; the exterior angle bisectors of the angles $CDA$ and $DAB$ intersect each other at $N$. Let $K_{1}$, $L_{1}$, $M_{1}$ and $N_{1}$ be the orthocenters of the triangles $ABK$, $BCL$, $CDM$ and $DAN$, respectively. Show that the quadrilateral $K_{1}L_{1}M_{1}N_{1}$ is a parallelogram.