In a triangle $ ABC$, where $ I$ is the incenter, and $ D,E,F$ the tangenty points of the incircle with $ BC,CA,AB$, and $ L \in BI \cap EF$, we have that $ BL \perp CL$. (A proof of this you can found in the begging of my proof of this problem: http://www.mathlinks.ro/Forum/viewtopic.php?t=132338) Now we just consider $ V,W$ as the intersection points of $ RS$ with the circumcircle of $ ABC$. Then $ \widehat{AV} = \widehat{BV}$ and $ \widehat{AW} = \widehat{CW}$, and $ \angle{MNU} = \frac {\widehat{CW} - \widehat{BV}}{2}$ , and $ \angle{MUN} = \angle{AUS} = \frac {\widehat{AW} - \widehat{AV}}{2}$, thus $ MUN$ is isosceles.

we see angles MUN= AUR=ARS-RAU=ASR-ACB=SNC so triangle MUN is isosceless q.e.d

How showed nicely Huyen Vu, this problem is truly generally and $M\equiv N$ iff $b^{2}+c^{2}=a(b+c)$.

easy to get <SRA=45;let <ACB=a; then <ABC=90-a;then <UNM=90-a-45=45-a;<UMB=90-2a=2<UNM;Hence :<UNM=<NUM......Q.E.D.

Attachments:

This problem is true generally, i.e. for any value of $ A$ . Prove easily that $ M\equiv N\Longleftrightarrow b^{2} + c^{2} = a(b + c)$ and $ MB > NB\Longleftrightarrow b^2 + c^2 > a(b + c)$ . In both cases, $ m(\widehat{MUN}) = m(\widehat {MNU}) = \frac 12\cdot|B - C|$ (see Huyền Vũ). Otherwise, if denote the diameter $ [AA']$ in the circumcircle $ C(O)$ of $ \triangle ABC$ and $ T\in AA'\cap RS$ , then observe that $ m(\widehat {AA'O})=\frac 12\cdot |B-C|$ and the quadrilaterals $ NTDA'$ , $ UASA'$ are cyclically a.s.o.

Por potencia de un punto AS=AR. ∆ASR triángulo rectángulo isósceles. Tenemos que ∠ASR=∠ARS=45° y por ser suplemento ∠BRN=135° Tenemos que ∠ABC=∠MAC=β por ángulo semiinscripto. En ∆ARU: β+135°+∠AUR=180° En ∆BRN: 135°+β+∠MUN=180° (1) Por opuesto por el vértice ∠AUR=∠MUN y por ello: β+135°+∠MUN=180° (2) Por (1) y (2): ∠BNR=∠MUN. ∴∆UMN es isósceles ∎