In making Euclidean constructions in geometry it is permitted to use a ruler and a pair of compasses. In the constructions considered in this question no compasses are permitted, but the ruler is assumed to have two parallel edges, which can be used for constructing two parallel lines through two given points whose distance is at least equal to the breadth of the rule. Then the distance between the parallel lines is equal to the breadth of the ruler. Carry through the following constructions with such a ruler. Construct: [b]a)[/b] The bisector of a given angle. [b]b)[/b] The midpoint of a given rectilinear line segment. [b]c)[/b] The center of a circle through three given non-collinear points. [b]d)[/b] A line through a given point parallel to a given line.

Let be a point $ P $ in the interior of a parallelogram $ ABCD $ such that $ \angle PAD=\angle PCD. $ Prove that the bisectors of $ \angle BAD $ and $ \angle BPD $ are parallel.

Two circles $C_{1}$ and $C_{2}$ are (externally or internally) tangent at a point $P$. The straight line $D$ is tangent at $A$ to one of the circles and cuts the other circle at the points $B$ and $C$. Prove that the straight line $PA$ is an interior or exterior bisector of the angle $\angle BPC$.

We draw two lines $(\ell_1) , (\ell_2)$ through the orthocenter $H$ of the triangle $ABC$ such that each one is dividing the triangle into two figures of equal area and equal perimeters. Find the angles of the triangle.

Bisector of side $BC$ intersects circumcircle of triangle $ABC$ in points $P$ and $Q$. Points $A$ and $P$ lie on the same side of line $BC$. Point $R$ is an orthogonal projection of point $P$ on line $AC$. Point $S$ is middle of line segment $AQ$. Show that points $A, B, R, S$ lie on one circle.

Let $ ABC $ be a triangle. The perpendicular in $ B $ of the bisector of the angle $ \angle ABC $ intersects the bisector of the angle $ \angle BAC $ in $ M. $ Show that $ MC $ is perpendicular to the bisector of $ \angle BCA. $

Let $ABCD$ be a trapezoid with bases $AB$ and $CD$. Bisectors of $AD$ and $BC$ intersect line segments $BC$ and $AD$ respectively in points $P$ and $Q$. Show that $\angle APD = \angle BQC$.

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 acute triangle $ABC$ bisector of angle $BAC$ intersects side $BC$ in point $D$. Bisector of line segment $AD$ intersects circumcircle of triangle $ABC$ in points $E$ and $F$. Show that circumcircle of triangle $DEF$ is tangent to line $BC$.

The incircle of a triangle $ABC$ has centre $I$ and touches sides $AB, BC, CA$ in points $C_1, A_1, B_1$ respectively. Denote by $L$ the foot of a bissector of angle $B$, and by $K$ the point of intersecting of lines $B_1I$ and $A_1C_1$. Prove that $KL\parallel BB_1$. [b]proposed by L. Emelyanov, S. Berlov[/b]

Prove that a line that divides a triangle into two polygons of equal area and equal perimeter passes through the center of the circle inscribed in the triangle. Prove an analogous property for a polygon that has an inscribed circle.

On the bisector $AL$ of triangle $ABC$ a point $K$ is chosen such that $\angle BKL=\angle KBL=30^{\circ}$. Lines $AB$ and $CK$ intersect at point $M$, lines $AC$ and $BK$ intersect at point $N$. FInd the measure of angle $\angle AMN$ [I]Proposed by D. Shiryaev, S. Berlov[/i]

Let $ABC$ be a triangle with bisectors $AA_1,BB_1, CC_1$ ($A_1 \in BC$, etc.) and $M$ their common point. Consider the triangles $MB_1A, MC_1A,MC_1B,MA_1B,MA_1C,MB_1C$, and their inscribed circles. Prove that if four of these six inscribed circles have equal radii, then $AB = BC = CA.$

The quadrilateral $ABCD$ is inscribed in the circle $o$. Bisectors of angles $DAB$ and $ABC$ intersect at point $P$, and bisectors of angles $BCD$ and $CDA$ intersect in point $Q$. Point $M$ is the center of this arc $BC$ of the circle $o$ which does not contain points $D$ and $A$. Point $N$ is the center of the arc $DA$ of the circle $o$, which does not contain points $B$ and $C$. Prove that the points $P$ and $Q$ lie on the line perpendicular to $MN$.

In the triangle $ABC$, angle $C$ is four times smaller than each of the other two angle The altitude $AK$ and the angle bisector $AL$ are drawn from the vertex of the angle $A$. It is known that the length of $AL$ is equal to $\ell$. Find the length of the segment $LK$. (Gryhoriy Filippovskyi)

Let $ABC$ be a triangle with bisectors $AA_1,BB_1, CC_1$ ($A_1 \in BC$, etc.) and $M$ their common point. Consider the triangles $MB_1A, MC_1A,MC_1B,MA_1B,MA_1C,MB_1C$, and their inscribed circles. Prove that if four of these six inscribed circles have equal radii, then $AB = BC = CA.$

The median $CM$ of the triangle $ABC$ is equal to the bisector $BL$, also $\angle BAC=2\angle ACM$. prove that the triangle is right.

Let $ABC$ be a triangle with $AB \neq AC$ and let $M$ be the middle of $BC$. The bisector of $\angle BAC$ intersects the line $BC$ in $Q$. Let $H$ be the foot of $A$ on $BC$. The perpendicular to $AQ$ passing through $A$ intersects the line $BC$ in $S$. Show that $MH \times QS=AB \times AC$.

Triangle $ABC$ with $AB = AC$ is inscribed in circle $o$. Circles $o_1$ and $o_2$ are internally tangent to circle $o$ in points $P$ and $Q$, respectively, and they are tangent to segments $AB$ and $AC$, respectively, and they are disjoint with the interior of triangle $ABC$. Let $m$ be a line tangent to circles $o_1$ and $o_2$, such that points $P$ and $Q$ lie on the opposite side than point $A$. Line $m$ cuts segments $AB$ and $AC$ in points $K$ and $L$, respectively. Prove, that intersection point of lines $PK$ and $QL$ lies on bisector of angle $BAC$.

In a triangle $ABC$, the bisectors of angles $A$ and $B$ meet the opposite sides of the triangle at points $D$ and $E$, respectively. A point $P$ is arbitrarily chosen on the line $DE$. Prove that the distance of $P$ from line $AB$ equals the sum or the difference of the distances of $P$ from lines $AC$ and $BC$.

In making Euclidean constructions in geometry it is permitted to use a ruler and a pair of compasses. In the constructions considered in this question no compasses are permitted, but the ruler is assumed to have two parallel edges, which can be used for constructing two parallel lines through two given points whose distance is at least equal to the breadth of the rule. Then the distance between the parallel lines is equal to the breadth of the ruler. Carry through the following constructions with such a ruler. Construct: [b]a)[/b] The bisector of a given angle. [b]b)[/b] The midpoint of a given rectilinear line segment. [b]c)[/b] The center of a circle through three given non-collinear points. [b]d)[/b] A line through a given point parallel to a given line.

Two circles $C_{1}$ and $C_{2}$ are (externally or internally) tangent at a point $P$. The straight line $D$ is tangent at $A$ to one of the circles and cuts the other circle at the points $B$ and $C$. Prove that the straight line $PA$ is an interior or exterior bisector of the angle $\angle BPC$.

Does there exist a nonisosceles triangle such that the altitude from one vertex, the bisectrix from the second one and the median from the third one are equal?

Since this is already 3 PM (GMT +7) in Jakarta, might as well post the problem here. Problem 1. Given an acute triangle $ABC$ and the point $D$ on segment $BC$. Circle $c_1$ passes through $A, D$ and its centre lies on $AC$. Whereas circle $c_2$ passes through $A, D$ and its centre lies on $AB$. Let $P \neq A$ be the intersection of $c_1$ with $AB$ and $Q \neq A$ be the intersection of $c_2$ with $AC$. Prove that $AD$ bisects $\angle{PDQ}$.

a) Does there exist a triangle in which the shortest median is longer that the longest bisectrix? b) Does there exist a triangle in which the shortest bisectrix is longer that the longest altitude?