Let $ABC$ be a triangle with $\angle{A} = 60^\circ$. The midperpendicular of segment $AB$ meets line $AC$ at point $C_1$. The midperpendicular of segment $AC$ meets line $AB$ at point $B_1$. Prove that line $B_1C_1$ touches the incircle of triangle $ABC$.

Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$. Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$. Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$.

In triangle $\vartriangle ABC$, we have marked points $A_1$ on side $BC, B_1$ on side $AC$, and $C_1$ on side $AB$ so that $AA_1$ is an altitude, $BB_1$ is a median, and $CC_1$ is an angle bisector. It is known that $\vartriangle A_1B_1C_1$ is equilateral. Prove that $\vartriangle ABC$ is equilateral too. (Note: A median connects a vertex of a triangle with the midpoint of the opposite side. Thus, for median $BB_1$ we know that $B_1$ is the midpoint of side $AC$ in $\vartriangle ABC$.)

On side $AB$ of a scalene triangle $ABC$ there are points $M$, $N$ such that $AN=AC$ and $BM=BC$. The line parallel to $BC$ through $M$ and the line parallel to $AC$ through $N$ intersect at $S$. Prove that $\measuredangle{CSM} = \measuredangle{CSN}$.

Let $ ABC$ be an acute triangle and $ CL$ be the angle bisector of $ \angle ACB$. The point $ P$ lies on the segment $CL$ such that $ \angle APB\equal{}\pi\minus{}\frac{_1}{^2}\angle ACB$. Let $ k_1$ and $ k_2$ be the circumcircles of the triangles $ APC$ and $ BPC$. $ BP\cap k_1\equal{}Q, AP\cap k_2\equal{}R$. The tangents to $ k_1$ at $ Q$ and $ k_2$ at $ B$ intersect at $ S$ and the tangents to $ k_1$ at $ A$ and $ k_2$ at $ R$ intersect at $ T$. Prove that $ AS\equal{}BT.$

Given a parallelogram $ABCD$ with $AB<BC$, show that the circumcircles of the triangles $APQ$ share a second common point (apart from $A$) as $P,Q$ move on the sides $BC,CD$ respectively s.t. $CP=CQ$.

Bisectrices $AA_1$ and $BB_1$ of triangle $ABC$ meet in $I$. Segments $A_1I$ and $B_1I$ are the bases of isosceles triangles with opposite vertices $A_2$ and $B_2$ lying on line $AB$. It is known that line $CI$ bisects segment $A_2B_2$. Is it true that triangle $ABC$ is isosceles?

Let $K$ be a point on the side $[AB]$, and $L$ be a point on the side $[BC]$ of the square $ABCD$. If $|AK|=3$, $|KB|=2$, and the distance of $K$ to the line $DL$ is $3$, what is $|BL|:|LC|$? $ \textbf{(A)}\ \frac78 \qquad\textbf{(B)}\ \frac{\sqrt 3}2 \qquad\textbf{(C)}\ \frac 87 \qquad\textbf{(D)}\ \frac 38 \qquad\textbf{(E)}\ \frac{\sqrt 2}2 $

Let $AL$ and $BK$ be the angle bisectors in a non-isosceles triangle $ABC,$ where $L$ lies on $BC$ and $K$ lies on $AC.$ The perpendicular bisector of $BK$ intersects the line $AL$ at $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK.$ Prove that $LN=NA.$

Let D be an arbitrary point inside $\Delta ABC$. Let $\Gamma$ be the circumcircle of $\Delta BCD$. The external angle bisector of $\angle ABC$ meets $\Gamma$ again at $E$. The external angle bisector of $\angle ACB$ meets $\Gamma$ again at $F$. The line $EF$ meets the extension of $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the circumcircles of $\Delta BFP$ and $\Delta CEQ$ always pass through the same fixed point regardless of the position of $D$. (Assume all the labelled points are distinct.)

Let $ABCD$ be a square and let $\Gamma$ be the circumcircle of $ABCD$. $M$ is a point of $\Gamma$ belonging to the arc $CD$ which doesn't contain $A$. $P$ and $R$ are respectively the intersection points of $(AM)$ with $[BD]$ and $[CD]$, $Q$ and $S$ are respectively the intersection points of $(BM)$ with $[AC]$ and $[DC]$. Prove that $(PS)$ and $(QR)$ are perpendicular.

$ D$ is a point on the edge $ BC$ of triangle $ ABC$ such that $ AD\equal{}\frac{BD^2}{AB\plus{}AD}\equal{}\frac{CD^2}{AC\plus{}AD}$. $ E$ is a point such that $ D$ is on $ [AE]$ and $ CD\equal{}\frac{DE^2}{CD\plus{}CE}$. Prove that $ AE\equal{}AB\plus{}AC$.

Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle{APD}$.

Let $ABC$ be an acute triangle with $| AB | > | AC |$. Let $D$ be the foot of the altitude from $A$ to $BC$, let $K$ be the intersection of $AD$ with the internal bisector of angle $B$, Let $M$ be the foot of the perpendicular from $B$ to $CK$ (it could be in the extension of segment $CK$) and$ N$ the intersection of $BM$ and $AK$ (it could be in the extension of the segments). Let $T$ be the intersection of$ AC$ with the line that passes through $N$ and parallel to $DM$. Prove that $BM$ is the internal bisector of the angle $\angle TBC$

In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ and $\theta$ is the measure of $\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals [asy] size(230); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(36), C=intersectionpoint(B--(9001,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b); draw(N--B--A--N--M--C--A^^B--M); markscalefactor=0.1; draw(rightanglemark(B,N,A)); pair point=N; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, S); label("$N$", N, dir(30)); label("$19$", (A+C)/2, dir(A--C)*dir(90)); label("$14$", (A+B)/2, dir(A--B)*dir(270)); [/asy] $\displaystyle \text{(A)} \ 2 \qquad \text{(B)} \ \frac{5}{2} \qquad \text{(C)} \ \frac{5}{2} - \sin \theta \qquad \text{(D)} \ \frac{5}{2} - \frac{1}{2} \sin \theta \qquad \text{(E)} \ \frac{5}{2} - \frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$