Let $D$ be the midpoint of the side $[BC]$ of the triangle $ABC$ with $AB \ne AC$ and $E$ the foot of the altitude from $BC$. If $P$ is the intersection point of the perpendicular bisector of the segment line $[DE]$ with the perpendicular from $D$ onto the the angle bisector of $BAC$, prove that $P$ is on the Euler circle of triangle $ABC$.

Let $H$ and $O$ be the orthocenter and circumcenter of an acute-angled triangle $ABC$, respectively. The perpendicular bisector of $BH$ meets $AB$ and $BC$ at points $A_1$ and $C_1$, respectively. Prove that $OB$ bisects the angle $A_1OC_1$.

In the right triangle $ABC,$ $CF$ is the altitude based on the hypotenuse $AB.$ The circle centered at $B$ and passing through $F$ and the circle with centre $A$ and the same radius intersect at a point of $CB.$ Determine the ratio $FB : BC.$

Let $ABC$ be a triangle with $\angle BAC \neq 90^{\circ}.$ Let $O$ be the circumcenter of the triangle $ABC$ and $\Gamma$ be the circumcircle of the triangle $BOC.$ Suppose that $\Gamma$ intersects the line segment $AB$ at $P$ different from $B$, and the line segment $AC$ at $Q$ different from $C.$ Let $ON$ be the diameter of the circle $\Gamma.$ Prove that the quadrilateral $APNQ$ is a parallelogram.

Chord $EF$ is the perpendicular bisector of chord $BC$, intersecting it in $M$. Between $B$ and $M$ point $U$ is taken, and $EU$ extended meets the circle in $A$. Then, for any selection of $U$, as described, triangle $EUM$ is similar to triangle: [asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pair F = (0, 1); pair M = midpoint(B--C); pair A = (-0.99, -0.141); pair U = intersectionpoints(A--E, B--C)[0]; draw(B--C); draw(F--E--A); draw(unitcircle); label("$B$", B, SW); label("$C$", C, SE); label("$A$", A, W); label("$E$", E, S); label("$U$", U, NE); label("$M$", M, NE); label("$F$", F, N); //Credit to MSTang for the asymptote [/asy] $\textbf{(A)}\ EFA \qquad \textbf{(B)}\ EFC \qquad \textbf{(C)}\ ABM \qquad \textbf{(D)}\ ABU \qquad \textbf{(E)}\ FMC$

The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$

Let $ABC$ be a triangle with $\angle A,\angle B,\angle C <120^{\circ}$, $T$ is its [i]Fermat-Torricelli[/i] point. Let $G$ be the centroid of $\triangle ABC$. Prove that: the distances from $G$ to the perpendicular bisectors of $TA, TB, TC$ are the same.

A regular polyhedron is a polyhedron that is convex and all of its faces are regular polygons. We call a regular polhedron a "[i]Choombam[/i]" iff none of its faces are triangles. a) prove that each choombam can be inscribed in a sphere. b) Prove that faces of each choombam are polygons of at most 3 kinds. (i.e. there is a set $\{m,n,q\}$ that each face of a choombam is $n$-gon or $m$-gon or $q$-gon.) c) Prove that there is only one choombam that its faces are pentagon and hexagon. (Soccer ball) [img]http://aycu08.webshots.com/image/5367/2001362702285797426_rs.jpg[/img] d) For $n>3$, a prism that its faces are 2 regular $n$-gons and $n$ squares, is a choombam. Prove that except these choombams there are finitely many choombams.

$\triangle{AOB}$ is a right triangle with $\angle{AOB}=90^{o}$. $C$ and $D$ are moving on $AO$ and $BO$ respectively such that $AC=BD$. Show that there is a fixed point $P$ through which the perpendicular bisector of $CD$ always passes.

Let $P$ be a point inside $\triangle ABC$. Let the perpendicular bisectors of $PA,PB,PC$ be $\ell_1,\ell_2,\ell_3$. Let $D =\ell_1 \cap \ell_2$ , $E=\ell_2 \cap \ell_3$, $F=\ell_3 \cap \ell_1$. If $A,B,C,D,E,F$ lie on a circle, prove that $C, P,D$ are collinear.

Let $I$ be the incenter of a triangle $ABC$. $D,E,F$ are the symmetric points of $I$ with respect to $BC,AC,AB$ respectively. Knowing that $D,E,F,B$ are concyclic,find all possible values of $\angle B$.

Let $ABC$ be an acute triangle and $\Gamma$ its circumcircle. Let $l$ be the line tangent to $\Gamma$ at $A$. Let $D$ and $E$ be the intersections of the circumference with center $B$ and radius $AB$ with lines $l$ and $AC$, respectively. Prove the orthocenter of $ABC$ lies on line $DE$.

Quadrangle $ABCD$ is inscribed in a circle with diameter $AC$. Points $K$ and $M$ are projections of vertices $A$ and $C$, respectively, onto line $BD$. A line parallel to $BC$ is drawn through point $K$ and intersecting $AC$ at point $P$. Prove that angle $KPM$ is a right angle.

Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$). [i]Greece[/i]

$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD$. The crease is $EF$, where $E$ is on $AB$ and $F$is on $CD$. The dimensions $AE=8$, $BE=17$, and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

Let $ K $ be the midpoint of the side $ AB $ of a given triangle $ ABC $. Let $ L $ and $ M$ be points on the sides $ AC $ and $ BC$, respectively, such that $ \angle CLK = \angle KMC $. Prove that the perpendiculars to the sides $ AB, AC, $ and $ BC $ passing through $ K,L, $ and $M$, respectively, are concurrent.

Let $ABC$ be an acute-angled, not equilateral triangle, where vertex $A$ lies on the perpendicular bisector of the segment $HO$, joining the orthocentre $H$ to the circumcentre $O$. Determine all possible values for the measure of angle $A$. (U.S.A. - 1989 IMO Shortlist)

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 $OI = IH$.

Lines $b$ and $c$ passing through vertices $B$ and $C$ of triangle $ABC$ are perpendicular to sideline $BC$. The perpendicular bisectors to $AC$ and $AB$ meet $b$ and $c$ at points $P$ and $Q$ respectively. Prove that line $PQ$ is perpendicular to median $AM$ of triangle $ABC$. (D. Prokopenko)

Line $\ell$ is perpendicular to one of the medians of the triangle. The median perpendiculars to the sides of this triangle intersect the line $\ell$ at three points. Prove that one of them is the midpoint of the segment formed by the other two.

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 $AB$.

In the trapezoid $ABCD$ we know that $CD \perp BC, $ and $CD \perp AD .$ Circle $w$ with diameter $AB$ intersects $AD$ in points $A$ and $P,$ tangent from $P$ to $w$ intersects $CD$ at $M.$ The second tangent from $M$ to $w$ touches $w$ at $Q.$ Prove that midpoint of $CD$ lies on $BQ.$

In a right-angled triangle $ABC$ ($\angle A = 90^o$), the perpendicular bisector of $BC$ intersects the line $AC$ in $K$ and the perpendicular bisector of $BK$ intersects the line $AB$ in $L$. If the line $CL$ be the internal bisector of angle $C$, find all possible values for angles $B$ and $C$. by Mahdi Etesami Fard

Cyclic pentagon $ ABCDE$ has a right angle $ \angle{ABC} \equal{} 90^{\circ}$ and side lengths $ AB \equal{} 15$ and $ BC \equal{} 20$. Supposing that $ AB \equal{} DE \equal{} EA$, find $ CD$.

A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas.