Let $N,K,L$ be points on the sides $\overline{AB}, \overline{BC}, \overline{CA}$ respectively. Suppose $AL=BK$ and $\overline{CN}$ is the internal bisector of angle $ACB$. Let $P$ be the intersection of lines $\overline{AK}$ and $\overline{BL}$ and let $I,J$ be the incenters of triangles $APL$ and $BPK$ respectively. Let $Q$ be the intersection of lines $\overline{IJ}$ and $\overline{CN}$. Prove that $IP=JQ$.

Let $ABC$ be an acute-angled triangle with circumcenter $O$ and incenter $I$. The midpoint of arc $BC$ of the circumcircle of $ABC$ not containing $A$ is denoted $S$. Points $E, F$ were chosen on line $OI$ for which $BE$ and $CF$ are both perpendicular to $OI$. Point $X$ was chosen so that $XE\perp AC$ and $XF\perp AB$. Point $Y$ was chosen so that $YE\perp SC$ and $YF\perp SB$. $D$ was chosen on $BC$ so that $DI\perp BC$. Prove that $X$, $Y$, and $D$ are collinear.

An acute-angled triangle $ABC$ is given. Points $A_1, B_1$ and $C_1$ are taken on its sides $BC, CA$ and $AB$, respectively, such that $\angle B_1A_1C_1 + 2 \angle BAC = 180^o$, $\angle A_1C_1B_1 + 2 \angle ACB = 180^o$, $\angle C_1B_1A_1 + 2 \angle CBA = 180^o$. Find the locus of the centers of the circles inscribed in triangles $A_1B_1C_1$ (all kinds of such triangles are considered).

In an isosceles trapezoid $ABCD$, the base $AB$ is twice as large as the base $CD$. Point $M$ is the midpoint of $AB$. It is known that the center of the circle inscribed in the triangle $MCB$ lies on the circle circumscribed around the triangle $MDC$. Find the angle $\angle MBC$. [img]https://cdn.artofproblemsolving.com/attachments/8/a/7af6a1d32c4e2affa49cb3eed9c10ba1e7ab71.png[/img]

Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies \[\angle PBA+\angle PCA = \angle PBC+\angle PCB.\] Show that $AP \geq AI$, and that equality holds if and only if $P=I$.

Given is the triangle $ABC$ such that $BC=13, CA=14, AB=15$ Prove that $B$, the incenter $J$ and the midpoints of $AB$ and $BC$ all lie on a circle

In right $ \triangle ABC$ with legs $ 5$ and $ 12$, arcs of circles are drawn, one with center $ A$ and radius $ 12$, the other with center $ B$ and radius $ 5$. They intersect the hypotenuse at $ M$ and $ N$. Then, $ MN$ has length: [asy]size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B), N=B+B.y*dir(B--A); real r=degrees(B); draw(A--B--C--cycle^^Arc(A,12,0,r)^^Arc(B,B.y,180+r,270)); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, dir(point--M)); label("$N$", N, dir(point--N)); label("$12$", (6,0), S); label("$5$", (12,3.5), E);[/asy] $ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac {13}{5} \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \frac {24}{5}$

In triangle $ABC$, $\angle C = 30^{\circ}$, $O$ and $I$ are the circumcenter and incenter respectively, Points $D \in AC$ and $E \in BC$, such that $AD = BE = AB$. Prove that $OI = DE$ and $OI \bot DE$.

The bisectors of the angles $A$ and $B$ of the triangle $ABC$ meet the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Assuming that $AE+BD=AB$, determine the angle $C$.

Let $ a\le b\le c$ be the sides of a right triangle, and let $ 2p$ be its perimeter. Show that \[ p(p \minus{} c) \equal{} (p \minus{} a)(p \minus{} b) \equal{} S, \] where $ S$ is the area of the triangle.

The figure shows a triangle, a circle circumscribed around it and the center of its inscribed circle. Using only one ruler (one-sided, without divisions), construct the center of the circumscribed circle.

Let $\triangle{ABC}$ be an acute-angled triangle and let $D$, $E$, $F$ be points on $BC$, $CA$, $AB$, respectively, such that \[\angle{AFE}=\angle{BFD}\mbox{,}\quad\angle{BDF}=\angle{CDE}\quad\mbox{and}\quad\angle{CED}=\angle{AEF}\mbox{.}\] Prove that $D$, $E$ and $F$ are the feet of the perpendiculars through $A$, $B$ and $C$ on $BC$, $CA$ and $AB$, respectively. [i](Swiss Mathematical Olympiad 2011, Final round, problem 2)[/i]

The circle of center $I$ is inscribed in the convex quadrilateral $ABCD$. Let $M$ and $N$ be points on the segments $AI$ and $CI$, respectively, such that $\angle MBN = \frac 12 \angle ABC$. Prove that $\angle MDN = \frac 12 \angle ADC$.

Let $ I$ be the incentre and $ O$ the circumcentre of a given acute triangle $ ABC$. The incircle is tangent to $ BC$ at $ D$. Assume that $ \angle B < \angle C$ and the segments $ AO$ and $ HD$ are parallel, where $H$ is the orthocentre of triangle $ABC$. Let the intersection of the line $ OD$ and $ AH$ be $ E$. If the midpoint of $ CI$ is $ F$, prove that $ E,F,I,O$ are concyclic.

Circles $\Omega $ and $\omega $ are tangent at a point $P$ ($\omega $ lies inside $\Omega $). A chord $AB$ of $\Omega $ is tangent to $\omega $ at $C;$ the line $PC$ meets $\Omega $ again at $Q.$ Chords $QR$ and $QS$ of $ \Omega $ are tangent to $\omega .$ Let $I,X,$ and $Y$ be the incenters of the triangles $APB,$ $ARB,$ and $ASB,$ respectively. Prove that $\angle PXI+\angle PYI=90^{\circ }.$

(A.Myakishev, 9--10) Given triangle $ ABC$. One of its excircles is tangent to the side $ BC$ at point $ A_1$ and to the extensions of two other sides. Another excircle is tangent to side $ AC$ at point $ B_1$. Segments $ AA_1$ and $ BB_1$ meet at point $ N$. Point $ P$ is chosen on the ray $ AA_1$ so that $ AP\equal{}NA_1$. Prove that $ P$ lies on the incircle.

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

Let $\triangle{ABC}$ be an acute-angled triangle and let $D$, $E$, $F$ be points on $BC$, $CA$, $AB$, respectively, such that \[\angle{AFE}=\angle{BFD}\mbox{,}\quad\angle{BDF}=\angle{CDE}\quad\mbox{and}\quad\angle{CED}=\angle{AEF}\mbox{.}\] Prove that $D$, $E$ and $F$ are the feet of the perpendiculars through $A$, $B$ and $C$ on $BC$, $CA$ and $AB$, respectively. [i](Swiss Mathematical Olympiad 2011, Final round, problem 2)[/i]

Circle $B$ has radius $6\sqrt{7}$. Circle $A$, centered at point $C$, has radius $\sqrt{7}$ and is contained in $B$. Let $L$ be the locus of centers $C$ such that there exists a point $D$ on the boundary of $B$ with the following property: if the tangents from $D$ to circle $A$ intersect circle $B$ again at $X$ and $Y$, then $XY$ is also tangent to $A$. Find the area contained by the boundary of $L$.

Let \( \omega \) be the circumcircle of triangle \( ABC \), where \( AB > AC \). Let \( N \) be the midpoint of arc \( \smile\!BAC \), and \( D \) a point on the circle \( \omega \) such that \( ND \perp AB \). Let \( I \) be the incenter of triangle \( ABC \). Reconstruct triangle \( ABC \), given the marked points \( A, D, \) and \( I \). Proposed by Oleksii Karlyuchenko and Hryhorii Filippovskyi

A non-isosceles triangle $\triangle ABC$ has its incircle tangent to $BC, CA, AB$ at points $D, E, F$. Let the incenter be $I$. Say $AD$ hits the incircle again at $G$, at let the tangent to the incircle at $G$ hit $AC$ at $H$. Let $IH \cap AD = K$, and let the foot of the perpendicular from $I$ to $AD$ be $L$. Prove that $IE \cdot IK= IC \cdot IL$.

Let $ABC$ a triangle with $AB<AC$ and let $I$ it´s incenter. Let $\Gamma$ the circumcircle of $\triangle BIC$. $AI$ intersect $\Gamma$ again in $P$. Let $Q$ a point in side $AC$ such that $AB=AQ$ and let $R$ a point in $AB$ with $B$ between $A$ and $R$ such that $AR=AC$. Prove that $IQPR$ is cyclic.

Let $ABC$ be a triangle inscribed in a circle ($\omega$) and $I$ is the incenter. Denote $D,E$ as the intersection of $AI,BI$ with ($\omega$). And $DE$ cuts $AC,BC$ at $F,G$ respectively. Let $P$ be a point such that $PF \parallel AD$ and $PG \parallel BE$. Suppose that the tangent lines of ($\omega$) at $A,B$ meet at $K$. Prove that three lines $AE,BD,KP$ are concurrent or parallel.

Let $I$ be the incenter of a non-equilateral triangle $ABC$ and $T_1$, $T_2$, and $T_3$ be the tangency points of the incircle with the sides $BC$, $CA$ and $AB$, respectively. Prove that the orthocenter of triangle $T_1T_2T_3$ lies on the line $OI$, where $O$ is the circumcenter of triangle $ABC$. [i]Proposed by Georgi Ganchev[/i]

Let $s$ be an arbitrary straight line passing through the incenter $I$ of the triangle $ABC$ . Line $s$ intersects lines $AB$ and $BC$ at points $D$ and $E$, respectively. Points $P$ and $Q$ are the centers of the circumscribed circles of triangles $DAI$ and $CEI$, respectively, and point $F$ is the second intersection point of these circles. Prove that the circumcircle of the triangle $PQF$ is always passes through a fixed point on the plane regardless of the position of the straight line $s$. (Matvii Kurskyi)