Hello. [asy][asy]import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.4121999989564498, xmax = 6.062806141586572, ymin = -2.2495502456549565, ymax = 3.111308491757533; /* image dimensions */ /* draw figures */ draw(circle((0.,0.), 1.)); draw((-1.2433917797462943,-0.9938732234903365)--(2.190773925818748,-1.0108286373975564)); draw((2.190773925818748,-1.0108286373975564)--(0.4603594250074033,0.9977392680600052)); draw((0.4603594250074033,0.9977392680600052)--(-0.8025010659795272,1.003974355172156)); draw((-0.8025010659795272,1.003974355172156)--(-1.2433917797462943,-0.9938732234903365)); draw((0.4603594250074033,0.9977392680600052)--(-1.2433917797462943,-0.9938732234903365)); draw(circle((0.46246977404039913,-0.21338472492368377), 0.7889011875711629)); draw((0.,0.)--(2.190773925818748,-1.0108286373975564)); draw((-0.1370059123494916,0.2994446669588031)--(0.4585748009132868,-1.0022762972804495)); draw((0.46246977404039913,-0.21338472492368377)--(-0.1370059123494916,0.2994446669588031)); draw((0.,0.)--(0.4603594250074033,0.9977392680600052)); draw((0.4603594250074033,0.9977392680600052)--(0.46246977404039913,-0.21338472492368377)); /* dots and labels */ dot((0.,0.),linewidth(3.pt) + dotstyle); label("$I$", (-0.1262251250275262,-0.05890032970234491), NE * labelscalefactor); dot((-0.004937212908886596,-0.9999878118900711),linewidth(3.pt) + dotstyle); label("$Q$", (-0.028269234476799462,-1.2254659353519066), NE * labelscalefactor); dot((-1.2433917797462943,-0.9938732234903365),linewidth(3.pt) + dotstyle); label("$A$", (-1.3818415402686597,-0.913788101781413), NE * labelscalefactor); dot((2.190773925818748,-1.0108286373975564),linewidth(3.pt) + dotstyle); label("$B$", (2.224716248189915,-0.9583135065771978), NE * labelscalefactor); dot((0.4603594250074033,0.9977392680600052),linewidth(3.pt) + dotstyle); label("$C$", (0.4971305421134621,1.054234790192275), NE * labelscalefactor); dot((-0.8025010659795272,1.003974355172156),linewidth(3.pt) + dotstyle); label("$D$", (-0.7673909540868284,1.054234790192275), NE * labelscalefactor); dot((0.46246977404039913,-0.21338472492368377),linewidth(3.pt) + dotstyle); label("$J$", (0.4437000563585202,-0.4061984871094663), NE * labelscalefactor); dot((0.4585748009132868,-1.0022762972804495),linewidth(3.pt) + dotstyle); label("$M$", (0.4704152992359912,-1.2076557734335926), NE * labelscalefactor); dot((-0.1370059123494916,0.2994446669588031),linewidth(3.pt) + dotstyle); label("$N$", (-0.25089625845572383,0.3863537182555031), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Suppose that $J$ is the incenter of $\triangle{ABC}$ and $I\equiv BJ\cap MN$.We will show that $I$ is the incenter of $ABCD$. Since $BI$ bisects $\angle{ABC}$,the above is equivalent to showing that $CI$ bisects $\angle{BCD}$.We will show that $CNIJ$ is cyclic. We have $\angle{JIN}=\angle{IBM}+\angle{IMB}=\frac{\angle{ABC}}{2}+180^{\circ}-\angle{AMN}=\frac{\angle{ABC}}{2}+180^{\circ}-\left(90^{\circ}-\frac{\angle{BAC}}{2}\right)=$ $=90^{\circ}+\frac{\angle{ABC}+\angle{BAC}}{2}=180^{\circ}-\frac{\angle{ACB}}{2}=180^{\circ}-\angle{NCJ}$.thus $CNIJ$ is indeed cyclic. Thus $\angle{CIJ}=90^{\circ}\Rightarrow \angle{ICB}=90^{\circ}-\frac{\angle{ABC}}{2}=\frac{\angle{BCD}}{2}$,and we are done.

it is well known that bk is perpendicular to ck after applying some obvious proprities( Mathot's theorem and that COD angle has 90 grades)..

If $O$ is the incenter of $\square ABCD$ and $I$ is the incenter of $\triangle ABC$ we have that $B,I,O$ are colinear. Now $\angle COD = \pi - \frac{\angle DCB + \angle CBA}{2} = \frac{\pi}{2} = \angle INC$. Therefore $CION$ is cyclic. Now: $$\angle INO = \angle ICO = \angle OCB - \angle ICB = \frac{\angle DCB - \angle ACB}{2} = \frac{\angle DCA}{2} = \frac{\angle BAC}{2} = \angle IAM = \angle INM$$ Hence $M,N,O$ are colinear.

Let $\Omega $ the incircle of $ABCD$ with incenter $O$ and let $\omega$ the incircle of $\triangle ABC$. Let $ T_{\infty} $ be the infinity point on $AB$ and $CD$ $\Longrightarrow$ the incircle of $\triangle T_{\infty}BC$ is $\Omega$ and the circumcircle of $\triangle T_{\infty}BC$ is the line $BC$ $\Longrightarrow$ since $\omega$ is tangent to $\odot (T_{\infty}BC)$, $CA$, $AB$ by Sawayama's theorem in $\odot (T_{\infty}BC)$ respect to cevian $CA$ we get $PQ$ through incenter of $\triangle T_{\infty}BC$ which is $O$ hence $M$, $N$ and $O$ are collinear.

This problem was proposed by Bulgaria.

We use the right angles on intouch chord lemma and angle chasing and we are done.

Claim 1: $IONC$ is cyclic with $\angle COI=90^{\circ}$ Proof: $\angle COI=180^{\circ}-(\angle OBC+\angle OCB)=180^{\circ}-\frac{180^{\circ}}{2}=90^{\circ}$ Claim 2: $\angle ICO=\angle IAM$ Proof: To make the work easies let's say that $\angle ICO=\alpha$ and $\angle ABI=CBI=\theta$ Then $\angle BCO=\frac{\angle BCD}{2}=90-\theta$ and $\angle BCI=\frac{\angle ACB}{2}=90-\alpha-\theta$ $\to$ $\angle ICO=\angle BCO-\angle BCI=\alpha$.DONE Back to the Problem: Since $\angle IMA=\angle INA=90^{\circ}$ we have $INAM$ is cyclic . Then we have $\angle INM=\angle IAM =\alpha (\star)$. And from Claim 1 and Claim 2 we have $\angle ICO=\angle INO =\angle IAM=\alpha(\star\star)$. Finally by combining $\star$ and $\star\star$ we get $\angle INO=\angle INM \ \to$ $N-O-M$ collinear.

Any analytic solution with coordinates or vectors? Please reply

Any ideas?

This problem is equivalent to the famous Iran incenter lemma!

Let $P$ be the point at infinity parallel to $AB, CD$. By Sawayama Theorem on $\triangle PBC$ and cevian $CA$, we are done. Corollary: if we define $M, N$ analogously for the $B$-excircle of $\triangle ABC$, the problem is still true.

bobthesmartypants wrote: Haha! You're so funny!

Let $O$ be the center of the circle, and $I$ the incenter of $\triangle{ABC}$. $CO \cap AB={X}$ $\angle{IOC}=180-(\angle{CBO}+ \angle{BCO})=90$ $\angle{AIC}=90+ \frac{\angle{B}}{2}, \angle{AXC}=\angle{BCX}=90- \frac{\angle{B}}{2} \rightarrow (A,X,C,Y)$ are concyclic. Thus $M,N,O$ is the Simson Line of $\triangle{AXC}$ wrt. $I$.

Let $I$ be the incenter of triangle $ABC$ and $O$ be the incenter of trapezoid $ABCD$. Note that $OC, OB$ are angle bisectors of $\angle C, \angle B$ respectively and $IA, IB, IC$ are angle bisectors of $\angle A, \angle B, \angle BCA$ respectively.Thus, $B, I, O$ are collinear as $I $and $O $ both lie on angle bisector of $\angle B$. Note that $\angle COI =180-(\angle OBC+\angle OCB)=180-(\frac{\angle B+\angle C}{2})=90$ as $AB \parallel CD$ so $\angle B+\angle C=180$. Thus, $\angle COI =\angle CNI=90$ so $CNOI$ is cyclic. Claim: $\angle ICO=\angle IMN$. $\angle OCD=\angle OCB $ $\angle BCI= \angle ICN$ Subtratcting the second equality from the first gives $\angle ICO=\angle ACD -\angle ICO$ or $\angle ICO=\frac{\angle ACD}{2}.$ But, $AB \parallel CD$ so $\angle ACD =\angle CAB=2\angle IAN.$ But, $INAM$ is cyclic so $\angle IAN= \angle IMN.$ So, $\angle ICO=\frac{\angle ACD}{2}=\angle IAN= \angle IMN,$ as desired. $\Box$ So, $\angle NOI=180-\angle ICN=180-\angle BCI= \angle OBC+\angle ICO + 90=\angle OBM +(\angle IMN +90)= \angle OBM +\angle OMB= 180-\angle IOM$, as desired. $ \blacksquare$

Let $I$ be incenter of $\Delta ABC$, and Let $I_X$ be incenter of $ABCD$, Hence Let $BI \cap MN =E \implies \angle BEC =90^{\circ} \text{ (by incircle chord lemma) } $, Since, $BI_X \text{ and }CI_X \text{ are also angle bisectors } \implies \boxed{ B-E-I_X}$ and we also know that, $\angle BI_XC =90^{\circ} \text{ (by angle chasing) } \implies \boxed{ E=I_X}$ and we also know that $E$ lies on $MN$, Hence, $\boxed{ I_X \in MN}$

neverlose wrote: A trapezoid $ABCD$ ($AB || CD$,$AB > CD$) is circumscribed.The incircle of the triangle $ABC$ touches the lines $AB$ and $AC$ at the points $M$ and $N$,respectively.Prove that the incenter of the trapezoid $ABCD$ lies on the line $MN$.