This one is easy. Move $M$ on $AI$. Then map $BM \mapsto BN$ is projective from pencil of lines through $B$ onto it self, since it is rotation for angle ${1\over 2}\angle ABC$ around $B$. There for map $M \mapsto N$ is projective form line $AI$ to line $CI$. Now this map induces projective map $\rho: DM \mapsto DN$ from pencil of lines through $D$ onto itself. Now, when $M= A$ then $N= I$, so $\angle MDN = {1\over 2} \angle ADC$ and when $M= I$ then $N= C$ so again $\angle MDN = {1\over 2} \angle ADC$, and when $M$ is incenter of triangle $ABD$, $N$ is incenter of triangle $BCD$, from where we get again $\angle MDN = {1\over 2} \angle ADC$. This implys that map $\rho$ is actualy rotation around $D$ for fixed angle ${1\over 2}\angle ADC$ since every projective map is exactly determined with 3 points.

my solution : let $\alpha$ be $\angle ABM$ and let $\beta$ be angle $CBN$. Then $\angle IBM=\beta$ and $\angle IBN=\alpha$. Then let $p,q,r,s$ be the angles in $D$. By the sine theorem we will get : $\displaystyle\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin BIA\sin BIC}{\sin A/2\sin C/2}$. And : $\displaystyle\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin p\sin s}{\sin r\sin q}\cdot\frac{\sin DIA\sin DIC}{\sin A/2\sin C/2}$. But $\sin BIA\sin BIC=\sin DIA\sin DIC$, because it will be the same with $\cos\frac{A-C}{2} +\cos(D+\frac{A+C}{2})=\cos\frac{C-A}{2} +\cos(B+\frac{A+C}{2})$, but this is true because $\cos x=\cos (-x)=\cos (2\pi-x)$. we will obtain that $\sin p\sin s=\sin r\sin q$ and because $p+q=r+s=D/2$, we transform into sum of cosines and we will obtain easy that $p=r$ and this solves the problem.

maky wrote: my solution : let $\alpha$ be $\angle ABM$ and let $\beta$ be angle $CBN$. Then $\angle IBM=\beta$ and $\angle IBN=\alpha$. Then let $p,q,r,s$ be the angles in $D$. By the sine theorem we will get : $\displaystyle\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin BIA\sin BIC}{\sin A/2\sin C/2}$. Can you show this step by step how to get Abdurashid

well, first : $\alpha=\angle ABM$ and $\beta=\angle CBN$ (notations). then, because $\angle BMN$ is half of $\angle B$, we get that $\angle MBI=\beta$ and $\angle NBI=\alpha$. also, $p=\angle CDN$, $q=\angle NDI$, $r=\angle IDM$ and $s=\angle MDA$. Then, sine theorem in $AMB$ and we get : $\frac{AM}{MB}=\frac{\sin\alpha}{\sin A/2}$. Sine theorem in $BMI$ : $\frac{MB}{MI}=\frac{\sin BIA}{\sin\beta}$. Sine theorem in $BNI$ : $\frac{BN}{IC}=\frac{\sin BIC}{\sin\alpha}$. Sine theorem in $BNC$ : $\frac{CN}{BN}=\frac{\sin\beta}{\sin C/2}$. We multiply this four relations and there goes the desired result : $\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin BIA\sin BIC}{\sin A/2\sin C/2}$ In a similar way, we get the other part : $\frac{AI}{MI}\cdot\frac{CI}{NI}=\frac{\sin p\sin s}{\sin r\sin q}\cdot\frac{\sin DIA\sin DIC}{\sin A/2\sin C/2}$. Now we show that $\sin BIA\sin BIC=\sin DIA\sin DIC$. (*) Let $A,B,C,D$ be the angles of $ABCD$. We get $\sin BIA=\sin\frac{A+B}{2}$, and similar relations. Then (*) becomes equivalent with : $\cos(\frac{A+B}{2}-\frac{C+B}{2})-\cos(\frac{B+A}{2}+\frac{B+C}{2})$ $=$ $\cos(\frac{A+D}{2}-\frac{C+D}{2})-\cos(\frac{D+A}{2}+\frac{D+C}{2})$. But we have $\cos\frac{A-C}{2}=\cos\frac{C-A}{2}$ and $\cos(B+\frac{A+C}{2})$ $=$ $\cos(2\pi-B-\frac{A+C}{2})=\cos(A+B+C+D-B-\frac{A+C}{2})=\cos(D+\frac{A+C}{2})$. But using the above facts, this becomes $\sin p\sin s=\sin q\sin r$. (**) Now let`s look at the angles $p,q,r,s$. We have $p+q=r+s=D/2$, and because $ABCD$ is convex, $D/2<\pi/2$, so $p+q<\pi/2$ and $r+s<\pi/2$. Also, $p,q,r,s$ are all smaller than $D/2$. Let $t=D/2$. Then $q=t-p$ and $s=t-r$. (**) means $\sin p\sin(t-r)=\sin r\sin(t-p)$, equivalent with $\cos(p-t+r) - \cos(p+t-r)=\cos(r-t+p)-\cos(r+t-p)$, which means $\cos(t+r-p)=\cos(t+p-r)$. We also have $0<t+r-p<2t<\pi$, and $0<t+p-r<\pi$. Then we must have $t+r-p=t+p-r$, which gives us $p=r$ and from here we get that $\MDC=q+r=q+p=D/2$.

Valentin Vornicu wrote: 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 $\angle MDN = \frac 12 \angle ADC$. Nice solution by inversion on the way Let $E$, $F$, $G$ and $H$ be the respective points of tangency of the circle, with sides $AB$, $BC$, $CD$, $DA$. Inversion, center $I$, radius $IE$. The images of $A$, $B$, $C$ and $D$ are $A'$, $B'$, $C'$ and $D'$, respective midpoints of segments $HE$, $EF$, $FG$, $GH$. Let $M'$ and $N'$ be the images of points $M$ and $N$. The inverted problem is formulated as follows: If $\angle B'M'I + \angle B'N'I = \angle B'A'I$ then $\angle D'M'I + \angle D'N'I = \angle D'A'I$. Now 1) $\angle B'A'I = \angle B'C'I$ properties of inversion, angle chasing or whatever 2) Hence $\angle M'B'A' = \angle B'N'I$, $\angle N'B'C' = \angle B'M'I$, hence triangles $M'B'A'$ and $N'B'C'$ are similar 3) Therefore $M'A' \cdot N'C' = A'B' \cdot B'C'$ 4) But $A'B' = C'D'$ and $B'C' = A'D'$ this is because all those points are midpoints of sides of a quadrilateral 5) Hence $M'A' \cdot N'C' = A'D' \cdot D'C'$ 6) Not hard to prove that $\angle M'A'D' = \angle N'C'D'$ hence triangles $M'A'D'$ and $N'C'D'$ are similar and hence result Daniel

I have two solution this beautiful problem: Proof 1: After some sine law , this problem is equivalent to $ \frac{AB \cdot BC}{AD \cdot DC}=\frac{BI^2}{DI^2} $ ,but it's well-known identity. Proof 2: Let the circles $ w_m=(M,r_n) $ and $ w_n=(N,r_n) $ such that $ w_m $ , is tangent to $ AB,AD $ and $ w_n $ tangent to $ BC,DC $ ,respectively.It's easy to see that $ 2 \angle MBN=ABC $ if and only if $ 2 \angle MDN=ADC $ if and only if $ B $ lies on one of two internal common tangent of $ w_m $ and $ w_n $.

My solution: Since $ AB-AD=CB-CD $ , so $ A, C $ lie on a hyperbola $ \Omega $ with focus $ B $ and $ D $ . Since $ \angle BAI=\angle IAD, \angle DCI=\angle ICB $ , so $ IA, IC $ is tangent to $ \Omega $ at $ A, C $, respectively . From $ \angle MBN=\frac{1}{2} \angle ABC $ we get $ MN $ is tangent to $ \Omega $ , so we get $ \angle MDN=\frac{1}{2} \angle ADC $ . Q.E.D